Why Does Monotone Convergence Theorem Confirm Integral Bounds?

[tjkubo]

Say f is a non-negative, integrable function over a measurable set E. Suppose
<br /> \int_{E_k} f\; dm \leq \epsilon<br />
for each positive integer k, where
<br /> E_k = E \cap [-k,k]<br />
Then, why is it true that
<br /> \int_E f\; dm \leq \epsilon \quad ?<br />

I know that
<br /> \bigcup_k E_k = E<br />
and intuitively it seems reasonable, but I don't know how to prove it.

 

[Eynstone]

Proceed by indirect method : if the integral over E is strictly > e ,then the integral over E_k should also exceed e for sufficiently large k.

 

[tjkubo]

Why is that so? Are you using some property of integrable functions I'm not seeing?

 

[spamiam]

I think you could also prove it this way. Given \epsilon &gt; 0, for each k \in \mathbb{Z}^+ \int_{E_k} f \; dm \leq \epsilon/2^k. Then
<br /> \int_E f \; dm = \int_{\bigcup_k E_k} f \; dm \leq \sum_{k=1}^\infty \int_{E_k} f \; dm \leq \sum_{k=1}^\infty \frac{\epsilon}{2^k} = \epsilon \frac{1}{2} \frac{1}{1 - \frac{1}{2}} = \epsilon \; .<br />

The only question I have is if it's legal to have an epsilon that depends on k, but I think it is since I think your given information held for all \epsilon &gt; 0 and all positive integers k.

 

[tjkubo]

Actually, the epsilon is a fixed number. That is why it's stumbling me. Sorry I wasn't clear.

 

[aesir]

define f_k(x)=\textbf{1}_{E_k}(x)f(x) where \textbf{1} is the characteristic function.
From the monotone convergence theorem you have
\epsilon \ge \lim_{k\to\infty} \int_{E_k} f \; dm = \lim_{k\to\infty} \int_E f_k \; dm = \int_E f \; dm