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Let $f$ be a integrable function on $E$. Show that for each $\epsilon > 0$, there is an $N \in \mathbb{N}$ for which $n \geq N$, then $|\int_{E_n} f|< \epsilon$ where $E_n=\{x \in E : |x| \geq n \}$.

I want to know if this is a proper way to prove this.

First note, by the continuity of measure, $\{E_n\}$ is a descending chain of sets. I was thinking this follows from the continuity of measure. But I'm not sure how to show that $|\int_{E_n}f|< \epsilon$. That is, I'm not sure how to show that I can make the measure over $E_n$ arbitrarily small.

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Consider the functions $z_n = f \cdot 1_{E_n}$. Then $|z_n| \leq |f|$, and $z_n(x) \to 0$ for each $x$. The dominated convergence theorem shows $\lim_{n\to \infty} \int |z_n| = 0$, and since $\int z_n = \int_{E_n} f$, the desired result follows (noting that $|\int g| \leq \int |g|$).

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