# Integrability of Derivative of a Continuous Function

Let $f$ be continuous on $[a,b]$ and has finite derivative a.e. on $[a,b]$. Let $f_n(x)=n[f(x+1/n)-f(x)]$ s.t. $f_n$ be uniformly integrable on $[a,b]$.

I want to show : $f'$ is Lebesgue integrable.

(I noted that $f_n\rightarrow f'$ pointwise.)

-
A pointwise limit of a uniformly integrable sequence is always integrable. –  Nate Eldredge Dec 10 '12 at 1:33

By the Mean Value Theorem, for each positive integer $n$ and for each $x \in [a,b]$, there exists $z_{n,x} \in (x, x+1)$ so that $$f_n(x)= f'(z_{n,x}).$$ It follows that $|f_n(x)| \leq M$ for all $n$ and for all $x\in [a,b]$.

Since $\{f_n\}$ converges pointwise to $f'$ on $[a,b]$, the Bounded Convergence Theorem yields $$\int_a^b f' = \lim_{n \to \infty } \int_a^b f_n$$ which shows that $f'$ is Lebesgue integrable.

-

This answer refers to a previous version of the (now edited) question, without assuming uniform integrability of $(f_n)$.

This is in general not true. A counterexample is $f(x) = x \sin \frac1x$ on $[0,1]$ (with $f(0)=0$). Then $f'(x) = \sin \frac1x - \frac1x \cos \frac1x$ is not Lebesgue-integrable over $[0,1]$, because $\int_0^1 |f'(x)| \, dx = \infty$ in this case.

You can even modify this example to be differentiable everywhere, using $g(x) = x^2 \sin \frac{1}{x^2}$. Then $g'(x) = 2x \sin \frac1{x^2} -\frac2x \cos\frac1{x^2}$ and again $\int_0^1 |g'(x)|\, dx = \infty$.

-
I think $f$ is not differentiable at $0$ –  Amr Dec 9 '12 at 23:53
Oh. Its OK because he said almost everywhere –  Amr Dec 9 '12 at 23:53
Is there a counerexample if a.e. is omitted ? –  Amr Dec 9 '12 at 23:54
@Lukas: I doubt that your example makes $f_n$ uniformly integrable. –  Avril B Dec 10 '12 at 0:06
@AvrilB: OK, now you just modified the assumptions in your question, then this counterexample does not work. Originally that assumption about uniform integrability was not there. –  Lukas Geyer Dec 10 '12 at 0:07