Prove that $\lim_{n \rightarrow \infty} \frac{1}{n}\int^n_0xf(x)dx = 0$

Let $f(x) \geq 0$ be continuous on the interval $[0, \infty)$, and suppose that $\int_0^\infty f(x)dx < \infty$. Prove that $\displaystyle \lim_{n \rightarrow \infty} \frac{1}{n}\int^n_0xf(x)dx = 0$

I want to use some version of dominated convergence theorem somewhere, and I have that the integral is equal to $\displaystyle \int_0^1nyf(ny)dy$ using change of variables. Some help would be great. Thanks.

2 Answers

Put $$g_n(x) = \frac{x}{n} f(x) \chi_{[0,n]}(x)$$ 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 $x$, and $g_n(x) \rightarrow 0$ pointwise. Therefore the dominated convergence theorem applies, and \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}

• @user46944 Seriously? – Did Aug 25 '14 at 17:54
• @Did That's pretty rude. – layman Aug 25 '14 at 17:54
• @Did It's rude to make someone feel as though their question is stupid. You can still accomplish your objective of making people inspect the problem to try and answer their own question without coming off as arrogant. – layman Aug 25 '14 at 17:59
• @user46944, as you can see from the votes, some here think your last comments are valuable, so don't feel pressured to delete them :) – Antonio Vargas Aug 25 '14 at 18:12
• @Did I'm not going to respond to this thread anymore because I think it is off topic from the question. Again, I'm sorry that you were offended. I understand your point completely about having people think about their questions before answering it for them. But I really hope you'll consider approaching this situation differently in the future. Instead of saying "Seriously?" you should be more gentle with the learner and ask leading questions that can help them arrive at the desired conclusion. – layman Aug 25 '14 at 18:12

Here's another, more elementary, approach that avoids measure theory:

Let $I_k = \int_{k}^{k+1} f(x) \, dx$; note that by the integrability assumption, S:= $\sum_{k\ge 0} I_k < \infty$. We may bound $$0\le \int_{k}^{k+1} x f(x) \, dx < (k+1) I_k$$ Thus, $$0\le \frac{1}{n}\int_{0}^{n} xf(x) \, dx < \frac{1}{n}\left(I_{0} + 2I_{1} + 3I_{2} + \cdots + n I_{n-1} \right)$$ Now, let $S_k = I_{0} + \cdots + I_{k-1} = \int_{0}^{k} f(x) \, dx$. We may rewrite the upper bound as $$\frac{1}{n} \left( S_n + (S_n - S_1) + (S_n - S_2) + \cdots + (S_n - S_{n-1}) \right)= S_n - \frac{S_1 + \cdots + S_{n-1}}{n}$$ Since $S_n \to S$ and $(S_1 + \cdots + S_{n-1})/n \to S$ as $n\to\infty$, it follows that the upper bound goes to zero as $n\to \infty$. By squeezing, we're done.