# Bounded Lebesgue integrals

1. Nov 1, 2011

### tjkubo

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

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

Last edited: Nov 1, 2011
2. Nov 1, 2011

### 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.

3. Nov 1, 2011

### tjkubo

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

4. Nov 1, 2011

### spamiam

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

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 $\epsilon > 0$ and all positive integers k.

5. Nov 1, 2011

### tjkubo

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

6. Nov 1, 2011

### 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
$$\epsilon \ge \lim_{k\to\infty} \int_{E_k} f \; dm = \lim_{k\to\infty} \int_E f_k \; dm = \int_E f \; dm$$