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Let $f$ be an extended real-valued $\mathcal{M}_{L}$-measurable function on $[0,\infty)$ such that $f$ is $\mu_L$-integrable on every finite subinterval of $[0,\infty)$, and $$ \lim_{x\rightarrow \infty}f(x)=c.$$ Let $a>0$. Show that $$\lim_{a\rightarrow \infty}\frac{1}{a}\int_{[0,a]}fd\mu_L=c$$

This is one of my analysis HW problems (9.35 in Yeh's Real Analysis, 3rd edition). I can solve it using some $\epsilon-\delta$ types argument(i.e. for large enough $x$, $c-\delta<f(x)<c+\delta$, use that to approximate the integral when $a$ is large enough).

However this chapter is about Lebesgue integral of measurable function and those convergence theorems(monotone convergence, dominated convergence, etc.). I wonder if there is a much better alternative solution for this problem using those theorems.

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    $\begingroup$ I wonder if one can justify that $\frac{1}{a}\int_{[0,a]}fd\mu_L = \int_0^1 f(ax) dx \to \int_0^1 cdx = c$. $\endgroup$ – Martin R Jan 29 '16 at 8:30
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If you assume that $f$ is bounded you may argue as follows: $${1\over a}\int_0^a f(x)\>dx=\int_0^1 f(a \>t)\>dt\ .$$ Now you can apply the dominated convergence theorem on the right hand side and obtain $$\lim_{a\to\infty}{1\over a}\int_0^a f(x)\>dx=\int_0^1 c\>dt=c\ .$$

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Your approach is the way to go. The point of this question I think is that infinity trumps all: even if $f$ never achieves the value $c$, since it is it's limit at infinity, the average value of $f$ on $[0,\infty)$ is still $c$. I could be wrong, but there is no obvious way to cast this as a direct application of one of the big convergence theorems.

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For $a$ large enough write

$$\int f1_{[0,a]}=\int f1_{[0,A]}+\int f1_{[A,a]},$$

where $A$ is s.t. $|f(x)-c|<\epsilon$ for some $\epsilon>0$ and $x\ge A$. The first integral on the RHS is finite by the given assumptions and the second integral is between $(c-\epsilon)(a-A)$ and $(c+\epsilon)(a+A)$. Combining these facts we can bound $\frac{1}{a}\int f1_{[0,a]}$. Then send $a\to \infty$ and $\epsilon\downarrow 0$.

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  • $\begingroup$ Still +1 for the solution, but that is the approach I use, and I'm wondering if there is any alternative using the related theorem in the chapter. $\endgroup$ – gamma Jan 29 '16 at 8:42
  • $\begingroup$ @frank000 I tend to agree with charlestoncrabb. $\endgroup$ – d.k.o. Jan 29 '16 at 8:49
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Probably the epsilon delta argument is the quickest approach (and the one you are meant to use). Anyway, try to see whether this works.

I assume f is bounded (for simplicity in f in (0,1)) although it is not an assumption (but it will start to be bounded at some point).

$\frac{1}{a} \int_0^a f dl=\int_0^{\infty} f \frac{1}{a} I_{(0,a)}dl=\int_0^{\infty} f dP_a$, where $P_a$ is the uniform probability measure on (0,a).

The last integral is equal to: $\int_0^1 P_a(f>u) du$.

Finally, using dominated convergence theorem and nothing that $P_a(f>u)$ converges to 1 if $u<c$ and 0 otherwise we get:

$\int_0^1 P_a(f>u) du \to c$

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