On the theorem of frullani

Theorem 3. Assuming the Continuum Hypothesis, there exists a nonmeasurable function /: R —> R such that, for every s e R,

8. 
   /(5 + 0 = fW
   

for almost every t G R.
Note that, for every s G R, there is a different set of measure zero Z₅, so that (8) does not imply that f is constant.
Proof. We shall construct f as the characteristic function of some set A c R. To construct A , we must first choose a certain Hamel basis of R over Q. Let Ж be the set of all compact subsets of R of positive Lebesgue measure. The cardinality of 5? is the same as that of R. So we can obtain a transfinite sequence IKy. a < tOj) with the members of . Now, by transfinite induc­tion, we define two transfinite sequences of real numbers {x_n: a < cu,) and (y_a: a < ct>j) in the way that follows.
First, since K_o is of positive measure, we can choose z₀, z'_o in K_o linearly independent over the field Q. We set x₀ = z₀, y₀ = 2z'_o.
Now, suppose we have chosen the sequences {x^: ft < a) and fi < a) so that the set B_n - {x^\f < a} U {y_p\P < a} is linearly independent over the field Q. As we assumed the Continuum Hypothesis, the set B_n is countable and, since K_(i has positive measure, we can select z_(t and z'_n G K_r such that B_n и {z_n, z'_t} is linearly independent over the field Q. Then we set = z_n and y_n = 2z'_n .
Now we complete the linearly independent set {xja < tu,} U {y_n|a < oj_}} to become an algebraic basis В of R over Q and we form a transfinite sequence (aja < co,) with the members of В.
We can define now the set A . Observe that every real number x / 0 can be expressed in just one way as a finite linear combination
n

8. 
   ⁼
   

k=l
where q_n G Q does not vanish and a_{ < a₂ < ■ ■ ■ < a_n . Let A be the set of those x G R such that q_ti is an integer.
Now we prove that, for every s G R, there exists Z_s of measure zero such that x G R\Z_s implies that x G A if and only if x + s & A. To see this, we note that, being
m
⁵ =
Z=1
the representation of 5 as a linear combination of elements of the basis В, every x expressed as in (9) satisfies, whenever a_n > , that x + s e A if
and only if x 6 A (we are supposing that > ft, if i < m ). So that we can choose Z_s as the linear span of the vectors of the set {afZ < ft_m}. Since Z_s is countable, it is a set of measure zero. This proves that the function f = X_A verifies, for every s e R,
/(s + r) = f(t) a.e.
Let us see now that f is non-Lebesgue measurable. We prove this assertion showing that, for every compact set К of positive measure, we have
(10) Х'пЛ/0, 7СП(К\Л)/0.
For, given such a compact К, there is an a < co, such that К — K_ti. Now, the expressions
, 1
Z — X , z — —y
(Y (Y ⁷ (Y 'y (Y
are the representations of the elements z ₍ and z'_n of К as linear combinations of elements of the basis В. So z_n & A and z'_n 0 A. This proves (10).

3. 
   Distributions and the Denjoy-Perron integral
   

It is usual to identify a locally integrable function /: R —> C with the distri­bution T_f: 2 C defined by ТД^) = f*™ f(t)
Nevertheless, there is a problem here: there exist locally integrable functions /: R —> C that admit a derivative f¹ (x) at all points x e R, without being f¹ locally integrable. In this situation we cannot identify f with DT_f.
However, if we use the Denjoy-Perron integral instead of the Lebesgue inte­gral, the problem vanishes. To prove this, we need a theorem of W. F. Pfeffer [22] on the Denjoy-Perron integral.
When we refer to the Denjoy-Perron integral we mean here the generalized Riemann integral as it is defined in R. M. McLeod [18]. This is justified, since both integrals are the same over compact intervals.