Riemann integral-Real analysis-infinity cases

Question

Let $f:\left[0,\infty\right] \to \mathbb{R}$ be a non-negative function such that $\int_{0}^{\infty}f(x)\,dx<\infty$ that $f$ is Riemann integrable on any finite interval $\left[0,R\right]$ and that $\lim_{R\to\infty}\int_{0}^{R}f(x)dx$ exists and is finite. Prove that the limit $\lim_{R\to\infty}\frac{1}{R}\int_{0}^{R}xf(x)dx$ exists and determine its value.

My approach is I first proved that product of two integrable functions is integrable using basic definitions of limit therefore the function $g(x)=xf(x)$ is integrable but then I tried integration by parts and I am stuck. The answer is probably zero but I could not exactly prove it using $\epsilon$ $\delta$ definition.

Answer 1

If you know about Cesàro means regarding to sequences, this is kind of the same idea.

Since $\int_0^{+\infty} f < +\infty$, we have $\int_t^{+\infty} f \to 0$ as $t \to +\infty$. Hence, if $\varepsilon > 0$, there exists $t_0 > 0$ such that $$t > t_0 \implies 0 \leq \int_t^{+\infty} f < \varepsilon.$$

However, since $f \geq 0$, \begin{align*}\frac1R\int_{0}^R x f(x) \mathrm dx = \frac1R \int_{0}^R \left(\int_0^x f(x) \mathrm dt \right)\mathrm dx &= \frac1R \int_{0}^R \left(\int_t^R f(x) \mathrm dx \right)\mathrm dt \\ &\leq \frac1R \int_{0}^R \left(\int_t^{+\infty} f \right)\mathrm dt. \end{align*}

Hence, \begin{align*}\frac1R\int_{0}^R x f(x) \mathrm dx &\leq \frac1R \int_{0}^{t_0} \left(\int_t^{+\infty} f \right)\mathrm dt + \frac1R \int_{t_0}^{R} \left(\int_t^{+\infty} f \right)\mathrm dt \\ &\leq \frac{t_0}{R} + \frac{R - t_0}{R}\varepsilon \end{align*}

This means that the $\overline{\lim}$ of $\frac1R\int_{0}^R x f(x) \mathrm dx$ can be arbitrarily small as $R \to +\infty$, i.e. its limit is $0$.

Here, I used the Fubini-Tonelli theorem, maybe you can avoid it, this is not exactly elementary...

You could also use a little $o$ summation theorem to shorten the proof.

Answer 2

Take $$g_n(x) = \frac{x}{n} f(x)$$ if $0\leq x\leq n$ and $g_n(x)=0$ for $x>n$. Then $$\frac{1}{n}\int_0^n xf(x) dx = \int_{0}^\infty g_n(x) dx$$ Also, $|g_n(x)| \leq |f(x)| = f(x)$ for all $x$, since $f$ is non negative. Clearly $g_n$ converges pointwise to $0$ to all $x\geq0$. We can apply the dominated convergence theorem, so $$\begin{align} \lim_{n \rightarrow \infty} \frac{1}{n}\int_0^n xf(x) dx &= \lim_{n \rightarrow \infty} \int_{0}^\infty g_n(x) dx\\ &= \int_{0}^{\infty} \lim_{n \rightarrow \infty}g_n(x) dx \\ &= 0 \end{align}$$

Answer 3

Sorry for double-posting. An alternative answer, without measure theory. Let $\varepsilon > 0$. There exists $R_0 > 0$ such that $0 \leq \int_{R_0}^{+\infty} f < \varepsilon$. Hence, if $R \geq R_0$,

\begin{align*} \frac1R \int_0^{R} xf(x) \mathrm dx &= \frac1R \underbrace{\int_0^{R_0}xf(x) \mathrm dx}_{=: \ M} + \int_{R_0}^{R} \frac xR f(x) \mathrm dx \\ &\leq \frac MR + \int_{R_0}^{R} f(x) \mathrm dx \\ &\leq \frac MR + \int_{R_0}^{+\infty} f(x) \mathrm dx \leq \frac MR + \varepsilon \underset{R \to +\infty}{\longrightarrow} \varepsilon. \end{align*}

This shows that the limit is indeed $0$.