Given that $f_n$ is an integrable function in $L(X,M,\mu)$ and that $$ \sum^\infty_1 \int |f_n| < \infty, $$ prove that $\sum f_n \to f\in L(X,M,\mu)$ a.e. Also that, $$ \int f = \sum^\infty_1 \int f_n. $$

I found this question in Elements of Integration by Bartles, and this is the proof I got so far. Could any of you guys look over it?

Let $g_n = \sum^n_1 |f_k|$. $g$ is measurable, and by the monotone convergence theorem, $$ \int \lim g_n = \sum^\infty_1 \int |f_n| < \infty. $$ Therefore, $g = \lim g_n$ is an integrable function in $L(X,M,\mu)$. Furthermore, since $\int g$ is finite, we know that $g(x) < \infty$ a.e. Therefore for a.e $x \in X$, $f_n(x)$ converges to a measurable function $f$. Since $f$ is dominated by $g$, we therefore conclude by the Lebesgue Dominated convergence theorem that $$ \int f = \sum^\infty_1 \int f_n. $$


1 Answer 1


It's basically fine, just a few glitches in the write-up.

At the beginning, you write "$g$ is measurable", but that should of course be $g_n$. (Typo)

Shortly before the end,

Therefore for a.e $x\in X$, $f_n(x)$ converges to a measurable function $f$.

but it's not the sequence $(f_n)$, it's the sequence of partial sums you're concerned with, let's call them

$$h_n = \sum_{k=1}^n f_k.$$

And finally,

Since $f$ is dominated by $g$

you don't need only $\lvert f\rvert \leqslant g$, you need the domination of the $h_n$.


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