This chapter is devoted to some elementary properties of monotone func- tions acting from R into R, and to some widely known examples of strange monotone functions. Let
f : R → R
be a partial function. We recall that f is said to be increasing (respectively, strictly increasing) if, for any two points x ∈ dom(f ) and y ∈ dom(f ), the relation x ≤ y (respectively, x < y) implies the relation f (x) ≤ f (y) (respectively, f (x) < f (y)).
Analogously the notion of a decreasing (respectively, strictly decreasing) partial function can be introduced. It is easy to see that f is increasing (strictly increasing) if and only if −f is decreasing (strictly decreasing).
A partial function from R into R is said to be monotone (respectively, strictly monotone) if it is either increasing or decreasing (respectively, either strictly increasing or strictly decreasing).
We shall consider below only increasing partial functions (this, of course, does not restrict the generality of our considerations).
Let f : R → R be an increasing bounded partial function and suppose that
dom(f ) 6= ∅. Fix a point t from dom(f ). For any x ≤ t, put
f∗(x) = inf {f (z) : z ∈ dom(f ), x ≤ z ≤ t} and, for any x ≥ t, put
f∗(x) = sup{f (z) : z ∈ dom(f ), t ≤ z ≤ x}.
It can easily be demonstrated that f∗ is an increasing bounded function
acting from R into R and extending f .
We thus see that every increasing bounded partial function admits an increasing bounded extension defined on the whole R. A similar argument shows that every increasing partial function f : R → R can be extended
to an increasing function defined on some subinterval of R. In view of this circumstance, we primarily will be dealing with increasing functions acting from a subinterval of R into R.
If f : R → R is an increasing function and x ∈ R, then there exist limits limy→x, y>xf (y) = f (x+),
limy→x, y<xf (y) = f (x−),
and we have the inequalities
f (x−) ≤ f (x) ≤ f (x+). Obviously f is continuous at x if and only if
f (x−) = f (x) = f (x+). Thus, x is a discontinuity point of f if and only if
f (x−) < f (x) ∨ f (x) < f (x+). More generally, let
g : R → R
be an arbitrary function and let x ∈ R. We say that x is a simple disconti- nuity point for g if there exist g(x−) and g(x+), but
g(x−) 6= g(x) ∨ g(x+) 6= g(x).
Evidently, if g is a monotone function, then all discontinuity points for g are its simple discontinuity points.
We have the following useful result.
Theorem 1. Let f be an arbitrary function acting from R into R. Then the set of all simple discontinuity points for f is at most countable.
Proof. Denote by E the set of all simple discontinuity points for f . Denote also:
E1= {x ∈ E : f (x−) < f (x+)},
E2= {x ∈ E : f (x−) > f (x+)},
E3= {x ∈ E : f (x) < f (x−) = f (x+)},
singular monotone functions 105
Then we can write
E = E1∪ E2∪ E3∪ E4,
ant it suffices to demonstrate that each set Ei (i = 1, 2, 3, 4) is at most
countable. Because the argument is similar for all Ei, we shall only show
that
card(E1) ≤ ω.
For this purpose, let us define a mapping
x → (p(x), q(x), r(x)) (x ∈ E1) such that p(x) ∈ Q, q(x) ∈ Q, r(x) ∈ Q, x ∈ ]q(x), r(x)[, (∀t ∈ ]q(x), x[)(f (t) < p(x)), (∀t ∈ ]x, r(x)[)(f (t) > p(x)).
The existence of such a mapping is obvious. Now, it is not hard to show that this mapping is injective. Indeed, suppose to the contrary that, for some distinct points x1and x2 from R, we have the equality
(p(x1), q(x1), r(x1)) = (p(x2), q(x2), r(x2)).
Without loss of generality we may assume that x1< x2. Choose any point
t from ]x1, x2[. Then we must have simultaneously
f (t) < p(x2) = p(x1),
f (t) > p(x1) = p(x2),
which is impossible. The contradiction obtained establishes that the map- ping
x → (p(x), q(x), r(x)) (x ∈ E1)
is injective, which, obviously, implies the relation card(E1) ≤ card(Q × Q × Q) = ω.
Theorem 1has thus been proved.
Exercise 1. Recall that a function f : R → R possesses the Darboux property if, for each subinterval [a, b] of R, the range of f contains the segment with the end-points f (a) and f (b).
Demonstrate that any function with the Darboux property has no simple discontinuity points. In particular, infer from this fact that if f is the derivative of some function acting from R into R, then f has no simple discontinuity points.
As a trivial consequence ofTheorem 1, we obtain that, for any monotone function f acting from R into R, the set D(f ) of all discontinuity points of f is at most countable.
Exercise 2. Let E = {xn : n ∈ N} be an arbitrary countable subset
of R and let {rn : n ∈ N} be a countable family of strictly positive real
numbers, such that
X
n∈N
rn < +∞.
For any x ∈ R, let us put
f (x) = X
n∈N (x)
rn,
where
N (x) = {n ∈ N : xn < x}.
In this way a certain function f acting from R into R is defined. Show that: (a) f is increasing;
(b) f is continuous at each point from R \ E; (c) for any natural index n, we have
f (xn+) − f (xn−) = rn;
in particular, f is discontinuous at each point of the given set E.
Deduce from this result that if E is everywhere dense in R (for example, if E = Q), then the function f constructed above has an everywhere dense set of its discontinuity points.
We now are going to present the classical Lebesgue theorem concerning the differentiation of monotone functions. For this purpose, we need three simple lemmas (cf. [154]).
First, let us recall the notion of a derived number for functions acting from R into R. Suppose that [a, b] is a segment of R and that f : [a, b] → R is a function. Let x ∈ [a, b]. We shall say that t ∈ R ∪ {−∞, +∞} is a derived number (or a Dini derived number) of f at x if there exists a sequence {xn : n ∈ N} of points from [a, b] tending to x, such that
(∀n ∈ N)(xn6= x), limn→+∞
f (xn) − f (x)
xn− x
singular monotone functions 107
In this case, we shall write
t = fD0 (x).
One more remark. For any two real numbers t1 and t2, it will be con-
venient to denote below by the symbol [t1, t2] the segment of R with the
end-points t1and t2. Thus, we do not assume in this notation that t1< t2.
Lemma 1. Let f : [a, b] → R be a strictly increasing function, let q be a positive real number and let X be a subset of [a, b] such that, for any point x ∈ X, there exists at least one derived number f0
D(x) ≤ q. Then we
have the inequality
λ∗(f (X)) ≤ qλ∗(X),
where λ is the standard Lebesgue measure on R and λ∗ denotes the outer measure associated with λ.
Proof. Fix an arbitrary ε > 0 and take an open subset G of [a, b] such that
X ⊂ G, λ(G) ≤ λ∗(X) + ε. Consider the family of segments
V = {[f (x), f (x + h)] : x ∈ X, h 6= 0, [x, x + h] ⊂ G, f (x + h) − f (x)
h ≤ q + ε}.
Clearly, this family forms a Vitali covering for the set f (X). Consequently, there exists a disjoint countable family
{[f (xn), f (xn+ hn)] : n ∈ N} ⊂ V
such that
λ(f (X) \ ∪{[f (xn), f (xn+ hn)] : n ∈ N}) = 0.
Note that, because our function f is strictly increasing, the countable family of segments
{[xn, xn+ hn] : n ∈ N}
is disjoint, too, and the union of this family is contained in G. So we can write λ∗(f (X)) ≤ X n∈N |f (xn+ hn) − f (xn)| ≤ (q + ε) X n∈N |hn|
≤ (q + ε)λ(G) ≤ (q + ε)(λ∗(X) + ε).
Taking account of the fact that ε > 0 was chosen arbitrarily, we conclude that
λ∗(f (X)) ≤ qλ∗(X), and the proof ofLemma 1is complete.
Lemma 2. Let f : [a, b] → R be a strictly increasing function, let q be a positive real number and let X be a subset of [a, b] such that, for any point x ∈ X, there exists at least one derived number fD0 (x) ≥ q. Then we have the inequality
λ∗(f (X)) ≥ qλ∗(X).
Proof. As we know, the set of all discontinuity points for f is at most countable. Taking this fact into account, we may assume without loss of generality that f is continuous at each point belonging to the given set X. Now, if q = 0, then there is nothing to prove. So let us suppose that q > 0. Pick an arbitrary ε > 0 for which q − ε > 0. There exists an open set G ⊂ R such that
f (X) ⊂ G, λ(G) ≤ λ∗(f (X)) + ε. Consider the family of segments
V = {[x, x + h] : x ∈ X, h 6= 0, [f (x), f (x + h)] ⊂ G, f (x + h) − f (x)
h ≥ q − ε}.
Obviously, this family forms a Vitali covering for the set X. Consequently, there exists a disjoint countable family
{[xn, xn+ hn] : n ∈ N} ⊂ V
for which we have
λ(X \ ∪{[xn, xn+ hn] : n ∈ N}) = 0.
Again, since our f is strictly increasing, the countable family of segments {[f (xn), f (xn+ hn)] : n ∈ N}
must be disjoint, too, and the union of this family is contained in G. Hence we may write (q − ε)λ∗(X) ≤ (q − ε)X n∈N |hn| ≤ X n∈N |f (xn+ hn) − f (xn)|
singular monotone functions 109
≤ λ(G) ≤ λ∗(f (X)) + ε.
Taking account of the fact that ε > 0 is arbitrarily small, we come to the desired inequality
qλ∗(X) ≤ λ∗(f (X)).
Lemma 2has thus been proved.
Lemma 3. Let f : [a, b] → R be a strictly increasing function and let X = {x ∈ [a, b] : there exist two distinct derived numbers of f at x}. Then X is a set of λ-measure zero.
Proof. For any two rational numbers p and q satisfying the inequalities 0 ≤ p < q,
let us denote
Xp,q= {x ∈ [a, b] : there exists a derived number of f at x
less than p, and there exists a derived number of f at x greater than q}. Clearly, we have
X = ∪{Xp,q : 0 ≤ p < q, p ∈ Q, q ∈ Q}.
So it suffices to show that each set Xp,q is of λ-measure zero. Indeed,
according toLemma 1, we may write
λ∗(f (Xp,q)) ≤ pλ∗(Xp,q).
At the same time, according toLemma 2, we have λ∗(f (Xp,q)) ≥ qλ∗(Xp,q).
These two inequalities yield
qλ∗(Xp,q) ≤ pλ∗(Xp,q)
or, equivalently,
0 ≤ (p − q)λ∗(Xp,q).
Because p − q < 0, we must have
This ends the proof of the lemma.
We are now ready to present the classical Lebesgue theorem on the differentiability (almost everywhere) of monotone functions.
Theorem 2. Let f : [a, b] → R be a monotone function. Then f is differentiable at almost all (with respect to λ) points of [a, b].
Proof. Obviously, we may suppose that f is increasing. Moreover, because the set of all differentiability points for f coincides with the set of all differentiability points for f1, where
f1(x) = f (x) + x (x ∈ [a, b]),
and f1is strictly increasing, we may assume without loss of generality that
our original function f is also strictly increasing. Now, in view ofLemma 3, it suffices only to demonstrate that the set
X = {x ∈ [a, b] : f or each n ∈ N, there exists a derived number fD0 (x) ≥ n}
is of λ-measure zero. But this follows directly fromLemma 2, because, in conformity with this lemma, we may write
nλ∗(X) ≤ λ∗(f (X)) ≤ f (b) − f (a),
λ∗(X) ≤ f (b) − f (a) n
for every natural number n ≥ 1, which immediately yields the required equality
λ∗(X) = 0. This completes the proof of Theorem 2.
It follows at once from this theorem that a nowhere differentiable real- valued function f defined on a segment [a, b] is simultaneously nowhere monotone on [a, b], i.e., there does not exist a nondegenerate subinterval of [a, b] on which f is monotone.
Exercise 3. Let f : R → R be a continuous function. Demonstrate that the following two assertions are equivalent:
(a) f is injective;
singular monotone functions 111
Show that there exist continuous functions g : R → R which cannot be represented in the form
g = g1+ g2,
where g1 and g2 are monotone functions acting from R into R.
On the other hand, by using the method of transfinite induction, prove that any function
h : R → R
is representable in the form h = h1+ h2, where both functions
h1: R → R, h2: R → R
are injective.
In this context, let us point out that if h is Lebesgue measurable, then both h1and h2 can be chosen to be Lebesgue measurable, too (see [139]).
Exercise 4. Let f : [a, b] → R be an increasing continuous function. Show that
Z b
a
f0(t)dt ≤ f (b) − f (a).
Give an example where this inequality is strict (cf. Theorem 4 below). In addition, demonstrate that if
Z b
a
f0(t)dt = f (b) − f (a),
then the function f is absolutely continuous on the whole segment [a, b]. Exercise 5. Let λ denote, as usual, the standard Lebesgue measure on R and let X be an arbitrary Lebesgue measure zero subset of R. Then there exists a sequence {Un : n ∈ N} of open subsets of R, such that
X ⊂ Un, λ(Un) < 1/2n (n ∈ N).
For any n ∈ N, let us define
fn(x) = λ(Un∩ ] − ∞, x]) (x ∈ R).
Then fn is increasing, continuous and
for all x ∈ R. Further, define fX(x) =
X
n∈N
fn(x) (x ∈ R).
Show that the function fX is increasing, continuous and, for any point
x ∈ X, the equality
limy→x, y6=x
fX(y) − fX(x)
y − x = +∞ holds true.
For our further considerations, we need the following useful result due to Fubini.
Theorem 3. Let {Fn : n ∈ N} be a sequence of positive (i.e., non-
negative) increasing functions given on a segment [a, b] ⊂ R, such that the seriesP
n∈NFn(x) converges for each point x ∈ [a, b] and
F (x) = X
n∈N
Fn(x) (x ∈ [a, b]).
Then, for almost all (with respect to the Lebesgue measure λ restricted to [a, b]) points x ∈ [a, b], the equality
F0(x) = X
n∈N
Fn0(x)
is satisfied.
Proof. Clearly, F is an increasing function on [a, b] and we may write F0(x) ≥ X
n∈N
Fn0(x)
for all those points x ∈ [a, b] at which the derivatives F0(x), F00(x), F10(x), ... , Fn0(x), ... exist. In view ofTheorem 2, the series of functionsP
n∈NFn0 is convergent
almost everywhere on [a, b]. Now, denote Sn(x) =
X
m≤n
singular monotone functions 113
and, for any natural number k, choose an index n(k) such that F (b) − Sn(k)(b) ≤ 1/2k.
Because all Fn are positive and increasing, we also have
0 ≤ F (x) − Sn(k)(x) ≤ 1/2k
for each x ∈ [a, b]. This implies that the series of increasing functions X
k∈N
(F (x) − Sn(k)(x))
converges uniformly on [a, b] to some increasing function. Applying Theo- rem 2once more, we easily infer that the series
X
k∈N
(F (x) − Sn(k)(x))0
converges at almost all points x ∈ [a, b]. From this fact we also claim that limk→+∞(F (x) − Sn(k)(x))0= 0
for almost all x ∈ [a, b], i.e.,
F0(x) = limk→+∞(F00(x) + F10(x) + ... + Fn(k)0 (x))
for almost all x ∈ [a, b]. But this immediately yields that F0= X
n∈N
Fn0
almost everywhere on [a, b]. The theorem has thus been proved.
The next exercise provides an application of this theorem to the differ- entiation of an indefinite Lebesgue integral.
Exercise 6. Let f be a positive lower semicontinuous function given on a segment [a, b]. We recall (seeExercise 18 from the Introduction) that f can be represented in the form
f = supn∈Nfn,
where all functions fn are positive, too, and continuous. Derive from this
fact that f can be also represented in the form f = X
n∈N
where all functions gn are positive and continuous.
Let now f be a positive, Lebesgue integrable, lower semicontinuous func- tion on [a, b] and let
F (x) = Z x
a
f (t)dt (x ∈ [a, b]).
Show, by applyingTheorem 3and the fact formulated above, that F0(x) = f (x)
for almost all (with respect to the Lebesgue measure) points x ∈ [a, b]. Let g be a positive, Lebesgue integrable function on [a, b]. Show that, for each ε > 0, there exists a lower semicontinuous function f on [a, b], such that g ≤ f and Z b a g(t)dt + ε > Z b a f (t)dt.
Deduce from this fact that there exists a sequence {fn : n ∈ N} of Lebesgue
integrable lower semicontinuous functions, such that: (a) fn+1≤ fn for any n ∈ N;
(b) g ≤ fn for any n ∈ N;
(c) limn→+∞fn(x) = g(x) for almost all points x ∈ [a, b].
In particular, we may write (f0− g) =
X
n∈N
(fn− fn+1)
almost everywhere on [a, b]. Observe that f0− g ≥ 0, fn− fn+1≥ 0 (n ∈ N). Putting F (x) = Z x a (f0− g)(t)dt (x ∈ [a, b]), Fn(x) = Z x a (fn− fn+1)(t)dt (x ∈ [a, b])
and applying Theorem 3 again, demonstrate that
( Z x
a
singular monotone functions 115
for almost all x ∈ [a, b]. Finally, prove the Lebesgue theorem stating that if h is an arbitrary real-valued Lebesgue integrable function on [a, b], then
( Z x
a
h(t)dt)0 = h(x)
for almost all x ∈ [a, b].
The exercise presented above shows us that the classical Lebesgue the- orem concerning the differentiation of a function H defined by
H(x) = Z x
a
h(t)dt (x ∈ [a, b]),
can be logically deduced fromTheorem 3. However, this approach has a weak side because it does not yield the description of the set of those points x ∈ [a, b] at which
H0(x) = h(x).
We now turn our attention to the construction of a strictly increasing function
g : R → R
whose derivative vanishes almost everywhere. Such a construction is essen- tially based on Theorem 3.
Let us recall that the first step of the construction of the Cantor set on R is that we remove from the segment [0, 1] the open interval ]1/3, 2/3[. Let us put at this step
f (x) = 0 (x ≤ 0), f (x) = 1 (x ≥ 1), f (x) = 1/2 (x ∈ ]1/3, 2/3[).
Now, suppose that on the n-th step of the construction we have already defined the function f for all those points which belong to the union of the removed (at this and earlier steps) intervals. Obviously, we obtain a finite family {[ai, bi] : 1 ≤ i ≤ m} of pairwise disjoint segments on [0, 1]. It is easy
to check that m = 2n, but we do not need this fact for our further purposes.
Pick any segment [ai, bi] from the above-mentioned family. Taking into
account the inductive assumption, we may put
f (x) =f (ai−) + f (bi+) 2
for all points
x ∈ ](2ai+ bi)/3, (2bi+ ai)/3[.
So we have defined our function f for all points belonging to the union of all intervals removed at the (n + 1)-th step. Continuing the process in this way, we will be able to construct f on the set R \ C, where C denotes the Cantor set. From the definition of f immediately follows that f is increasing and continuous on its domain. Moreover, it is easily seen that f can be uniquely extended to an increasing continuous function
f : R → [0, 1].
Because f is constant on each removed interval, we obviously have f0(x) = 0 (x ∈ R \ C),
i.e., the derivative of f vanishes almost everywhere on R.
Thus, we have shown that there exists a nonconstant increasing bounded continuous function f from R into R, whose derivative is zero almost ev- erywhere. Now, let p and q be any two points of R such that p < q. Because f is not constant, there are some points x and y from R such that f (x) < f (y). Evidently, x < y and there exists a homothety (or translation) h of the plane R2, for which
h((x, 0)) = (p, 0), h((y, 0)) = (q, 0).
Let f∗ denote the function from R into R, whose graph coincides with the image of the graph of f with respect to h. Then we may assert that f∗ is also an increasing bounded continuous function, whose derivative vanishes almost everywhere, and
f∗(p) < f∗(q).
In virtue of the remarks made above, we can formulate and prove the following classical result concerning the existence of strictly increasing con- tinuous singular functions.
Theorem 4. There exists a function g : R → R satisfying these three conditions:
(1) g is continuous and strictly increasing; (2) (∀x ∈ R)(0 ≤ g(x) ≤ 1);
singular monotone functions 117
Proof. Let {(pn, qn) : n ∈ N} denote the countable family of all
pairs of rational numbers, such that pn < qn. According to the argument
presented above, for each natural index n, there exists a function gn : R → [0, 1]
such that:
(a) gn is continuous and increasing;
(b) 0 ≤ gn(x) ≤ 1/2n+1for all x ∈ R;
(c) the inequality gn(pn) < gn(qn) holds true;
(d) the derivative of gn vanishes almost everywhere.
It follows from (b) that the seriesP
n∈Ngn is uniformly convergent. So
we may consider the function
g = X
n∈N
gn
that is continuous and increasing because of (a). Evidently, 0 ≤ g(x) ≤ 1 (x ∈ R).
In accordance with (c), we also have
g(pn) < g(qn) (n ∈ N),
which immediately implies that g is a strictly increasing function. Finally, taking into account condition (d) and applying Theorem 3, we conclude that the derivative of g equals zero almost everywhere on R.
Exercise 7. Let f be any continuous increasing function acting from R into R, whose derivative is equal to zero almost everywhere on R. For each half-open subinterval [a, b[ of R, let us put
µ([a, b[) = f (b) − f (a).
Show that µ can be uniquely extended to a σ-finite Borel measure on R (denoted by the same symbol µ) that is diffused (i.e., vanishes on all one- element subsets of R) and is singular with respect to the standard Lebesgue measure λ. The latter means that there exists a Borel subset X of R for which we have
λ(X) = 0, µ(R \ X) = 0.
Formulate the converse assertion and prove it by utilizing the Vitali covering theorem.
Exercise 8. Demonstrate that:
(a) an increasing function f : R → R is continuous from the right if and only if it is upper semicontinuous;
(b) an increasing function f : R → R is continuous from the left if and only if it is lower semicontinuous.
Starting with these facts, give an example of a monotone function g : R → R
that is upper (lower) semicontinuous and whose discontinuity points form an everywhere dense subset of R (cf. Exercise 2 of this chapter).
Exercise 9. Let f : R → R be an arbitrary bounded from above function. We define a function
f∗: R → R by putting
f∗(x) = supy<xf (y) (x ∈ R).
Verify that f∗ is increasing and lower semicontinuous.
In addition, suppose that the original function f is increasing. Show that f and f∗ have the same set of continuity points. In general, f∗ does not coincide with f . Check that f∗ coincides with f if and only if f is continuous from the left.
Exercise 10. Let {tn: n < ω} be an arbitrary sequence of real numbers.
Prove that at least one of the following three assertions is valid: (a) {tn: n < ω} contains an infinite strictly increasing subsequence;
(b) {tn: n < ω} contains an infinite strictly decreasing subsequence;
(c) {tn : n < ω} contains infinitely many terms which are equal to each
other.
Give a straightforward proof of this result. On the other hand, show that the same result is a direct consequence of the Ramsey combinatorial theorem for countable graphs [170].
Conclude from the said above that any partial function f : R → R
satisfying the relation
card(dom(f )) ≥ ω
either is strictly monotone on some infinite subset of R or is constant on some infinite subset of R.
singular monotone functions 119
The last result cannot be extended to uncountable subsets of dom(f ) (within the theory ZFC). Indeed, we shall see in our further considerations that (under CH) there exists a function
g : R → R,
which is not monotone on any uncountable subset of R. The construction of such a function is very similar to the classical construction of a Sierpi´nski- Zygmund function (cf. Chapter 7) or to the classical construction of a Luzin set on R (cf. Chapter 10).
Exercise 11. Let f : R → R be an arbitrary continuous function. Prove that there exists a nonempty perfect set P ⊂ R such that the restric- tion f |P is monotone on P .
This can be done by using the following (fairly standard) argument. Suppose that there exists no nondegenerate subinterval of R on which f is decreasing. In this case, construct by recursion a dyadic system
(Ti1i2...ik)i1∈{0,1},i2∈{0,1},...,ik∈{0,1},k≥0
of nondegenerate compact subintervals of R satisfying the relations: (a) Ti1i2...ikik+1⊂ Ti1i2...ik;
(b) Ti1i2...ik0∩ Ti1i2...ik1= ∅;
(c) diam(Ti1i2...ik) < 1/2 k;
(d) if (i1, i2, ..., ik) ≺ (j1, j2, ..., jk), then x < y and f (x) < f (y) for
all points x ∈ Ti1i2...ik and y ∈ Tj1j2...jk (here denotes the standard
lexicographical ordering on the set of all k-sequences whose terms belong to {0, 1}).
Finally, put
P = \
k≥0
(∪{Ti1i2...ik : i1, i2, ..., ik ∈ {0, 1}})
and verify that f is increasing on the nonempty perfect set P . Establish the same result for those functions
f : R → R,
which are Lebesgue measurable or possess the Baire property (reduce this more general situation to the case of a continuous real-valued function given on a nonempty perfect subset of R).
Exercise 12. Give an example of a Peano type mapping f = (f1, f2) : [0, 1] → [0, 1]2
such that both coordinate functions f1 and f2 are differentiable at almost
all points of [0, 1] (in the sense of λ) and the equalities f10 = f20 = 0
hold almost everywhere on [0, 1] (in the same sense).