# Continuous, integrable function with integrable derivative in $\mathbb{R}$

While studying for my exams, I came across this question and I'm trying to think about an intelligent way to solve it (the context is Lebesgue integration):

Let $f:\mathbb{R} \to \mathbb{R}$, a continuous function (please see note below) on every bounded interval. Show that if $f$ and $f'$ are integrable in $\mathbb{R}$ (meaning $\displaystyle \int_{\mathbb{R}} f(x) dx < \infty$ and $\displaystyle\int_{\mathbb{R}} f'(x) dx < \infty$), then:

$$\lim_{x \to \infty} f(x) = \lim_{x \to -\infty} f(x) = 0$$

and:

$$\int_{-\infty}^{\infty} f'(x)dx = 0$$

Definitions:

I Don't know how it's called in English: for every $\epsilon > 0$ there is a $\delta > 0$ such that for every ${[x_i, y_i]_{i=1}^{n}}$, if $\sum (y_i - x_i) < \delta$, then $\sum |f(y_i) - f(x_i)| < \epsilon$

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I believe that you are thinking of abslute continuity. Is that correct? –  JavaMan Feb 24 '12 at 16:28
Yep. You provide us the definition of absolute continuity. –  leo Feb 24 '12 at 16:36
@JavaMan, you forgot the $in \int_{\mathbb{R}} f'(x) dx < \infty). – leo Feb 24 '12 at 16:38 @leo: Thanks for pointing that out. It seems someone else fixed it in the meantime. – JavaMan Feb 24 '12 at 16:43 @Hila, please note that$f(x)$don't depends on$n$, so $$\lim_{n\to\infty}f(x)=f(x).$$ Then that you are asking for is:$f$integrable and$f'$integrable, implies,$f$and$f'$are equal to the constant function$0$and$\int_{-\infty}^\infty f'=0$. In that case, there is lots of counterexamples – leo Feb 24 '12 at 16:49 ## 1 Answer I presume you know that under these assumptions, the fundamental theorem of calculus (FTC) holds and we have$f(b) - f(a) = \int_a^b f'(x) dx$. First try showing that$f(x)$is Cauchy as$x \to \infty$: if$a,b$are very large then$|f(b) - f(a)|$must be very small. (Use FTC and the fact that$f'$is integrable.) This implies that$\lim_{x \to \infty} f(x)$exists. Show that the integrability of$f$means the limit must be 0. The argument as$x \to -\infty$is identical. Finally, note that$\int_{-m}^m f'(x)dx = f(m) - f(-m) \to 0$as$m \to \infty$. Use dominated convergence to conclude$\int_\mathbb{R} f'(x)dx =0$. - And then, why is$\int_{\mathbb{R}} f' = 0\$? –  Hila Feb 24 '12 at 17:57
@Hila: I edited to add this. –  Nate Eldredge Feb 24 '12 at 18:10