Lemma: If $f$ is a nonnegative measurable function and $\int f<\infty$, then $\int f = \lim_{n\to \infty}\int_{[-n,n]} f$.
Proof: We know that $\int_{[-n,n]}f = \int f \cdot \chi_{[-n,n]}$. If $n < m$, then $f \cdot \chi_{[-n,n]} \leq f \cdot \chi_{[-m,m]} \leq f$. So the limit exists and $\lim_{N\to \infty} \int_{[-N,N]} f\leq \int f$.
Now we want to show the reverse. Suppose $\epsilon > 0$. There exists a simple integrable function $g$ such that $0 \leq g \leq f$, and $\int f \leq \int g + \epsilon$. Let $n$ be sufficiently large such that $g = 0$ outside of $[-n,n]$. Then, $g = g \cdot \chi_{[-n,n]}$, so $\int g = \int_{[-n,n]} g$. Now $g\leq f$ implies $g \cdot \chi_{[-n,n]} \leq f \cdot \chi_{[-n,n]}$. Therefore, $\int f \leq \int g + \epsilon = \int_{[-n,n]}g + \epsilon \leq \int_{[-n,n]}f + \epsilon$ for $n > N \in \mathbb{N}$. So $\int f \leq \lim_{n\to \infty} \int_{[-n,n]} f + \epsilon$. Now letting $\epsilon \to 0$, we obtain $\int f \leq \lim_{n\to \infty} \int_{[-n,n]} f $, and we are done.
There is a problem in this proof, but I can't seem to find/correct it. Can anyone help?