Show that $\lim_{x\to\infty} \frac{1}{x} \int_0^x f(x) \ dx = \gamma$.

The Assignment

Let $f: [0,\infty) \rightarrow \mathbb{R}$ be a function which is integrable on the intervall $[0,x] \ \forall x > 0$ and $\lim_{x\to\infty} f(x) = \gamma \in \mathbb{R}$.

Show that $$\lim_{x\to\infty} \frac{1}{x} \int_0^x f(t) \ dt = \gamma$$

My first try failed pretty hard because I forgot that not every integrable function does have to have a antiderivative / indefinite integral.

I thought maybe splitting the integral into two parts using the convergence of $f$ might help, but I don't know where to go from there.

I'd appreciate any help.

• Be careful here: you use $x$ both as a quantity outside (and upperbound) of the integral, and as "dummy variable" for the integrand. – Clement C. Apr 28 '14 at 14:04
• What does $\lim_{x \to \infty} f(x)$ mean if $f$ is only defined on $[a,b]$? – Najib Idrissi Apr 28 '14 at 14:05
• @NajibIdrissi There were quite a lot of mistakes in the initial question. I corrected it. – Nhat Apr 28 '14 at 16:19

Hint: do the change of variable $u=\frac{t}{x}$, so that $$\frac{1}{x} \int_0^x f(t) dt = \int_0^1 f(ux) du.$$
• Is $\lim_{x\to\infty} \int_0^1 f(ux) du = \int_0^1 \lim_{x\to\infty} f(ux) du$ ? If so, why? Edit: I just realized, we cannot use substitution since $f$ is not necessarily continuous. – Nhat Apr 28 '14 at 16:59
• Can you explain why it still works? One of the main conditions of the substitution method is that $f$ should be continuous. – Nhat Apr 28 '14 at 22:03
• For Lebesgue's integral, you only need $f$ to be integrable to get $$\int_{\varphi(\Omega)} f(y)dy=\int_\Omega f(\varphi(x))\lvert \varphi^\prime(x)\rvert dx$$ as long as $\varphi$ is a diffeomorphism. $f$'s continuity is not necessary. – Clement C. Apr 28 '14 at 22:28
• I am only familiar with the Riemann integral, but that's good to know though. I think I've got a solution now though, constructing two continuous functions $g,h: g < f < h$ both having the limit $=\gamma$ and using the sandwich theorem. I'll post it later, so we can close this thread. – Nhat Apr 28 '14 at 22:37