Suppose $\{f_n \}\subset L^{+}$, $f_n\rightarrow f$ pointwise, and $\int f = \lim\int f_n < \infty$. Then $\int_{E}f = \lim\int_{E}f_n$ for all $E\in M$. However, this need not be true if $\int f = \lim\int f_n = \infty$

Proof: Let $E\in M$, by Fatou's Lemma $$\int_{E}f = \int f 1_{E} = \int\lim_{n\rightarrow \infty}\inf f_n 1_{E} \leq \lim_{n\rightarrow \infty}\inf\int f_n 1_{E} = \lim_{n\rightarrow \infty} \inf\int_{E}f_n$$ Similarly, $$\int_{E^{c}}f\leq \lim_{n\rightarrow \infty}\inf\int_{E^{c}}f_n$$ Note, $f = \int f 1_{E} + f 1_{E^{c}}$ and $f_n = f_n 1_{E} + f_n 1_{E^c}$ for all $n\in\mathbb{N}$. This implies, $\int f_{E^c} = \int f - \int_{E} f$ and (for $n$ large) $\int_{E^c}f_n = \int f_n - \int_{E} f_n$. Therefore, $$\int f - \int_{E}f = \int_{E^c}f\leq \lim_{n\rightarrow \infty}\int_{E^c}f_n = \lim_{n\rightarrow\infty}\left(\int f_n - \int_{E} f_n\right) = \int f - \lim_{n\rightarrow \infty}\int_{E}f_n$$ Cancelling $\int f$ from both sides (which is allowed since $\int f = \lim\int f_n < \infty$) we get $$\int_{E} f = \lim_{n\rightarrow \infty}\int_{E}f_n$$

I have not completed the second part yet, I just wanted to make sure this is correct, any suggestions is greatly appreciated especially for the second part.


For the second part consider $f = \chi_{(0,\infty)}$ and $$f_n = \chi_{(0,\infty)} + n^2\chi_{(-\frac{1}{n},0]}.$$

Then $f_n \to f$ pointwise, $\int_{\mathbb{R}}f = \infty = \lim_{n\to \infty}\int_{\mathbb{R}}f_n$, but for $E = (-\infty,0)$

$$ \int_E f = 0 \quad \text{and}\ \int_E f_n = n^2\int_{-\frac 1n}^01 = n. $$

Then $$\int_E f = 0 \neq \infty = \lim_n\int_E f_n.$$

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