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I have seen this question here but I didnt find a comprehensible answer so ill try my luck.

Let $(f_n)$ be a sequence of almost everywhere integrable functions defined on X with $$\sum_{n=0}^\infty \int_X |f_n| \, d \mu \lt \infty$$

Show that $$f(x) := \sum_{n=0}^\infty f_n(x)$$

$1)$ converges for almost all $x$

$2)$ the limit function $f$ is integrable

$3) \int_X f \, d\mu = \sum_{n=0}^\infty \int_Xf_n \, d\mu$

My try:

$1)$ Define $g_N := \sum_{n=0}^N|f_n|.$ It follows that $g_N$ is a monotone increasing function so I can use the monotone convergence theorem.

It follows that : $$\lim_{N \rightarrow \infty}\int_X g_N \, d \mu = \int_X \lim_{N \rightarrow \infty} g_N \, d \mu $$

$\Longrightarrow \sum_{n=0}^\infty|f_n| \lt \infty$ so the sum converges for almost all $x$

$2)$ Since all $f_n $ are measurable and we showed that $f = \lim f_n$ is finite it follows that $f$ is integrable

$3)$ I would say it strictly follows from dominated convergence theorem with $1)$ and $2)$.

Any help here is appreciated thanks

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2 Answers 2

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Let $$ F_n= \sum_{k=0}^{n}f_k(x); \quad F(x)= \sum_{k=0}^{\infty}f_k(x) $$ Clearly, $F_n\in L^1$. Further $$ |F_n|\leq \sum_{k=0}^{n} |f_k(x)|\leq \sum_{k=0}^{\infty} |f_k(x)|\in L^1\tag{1} $$ since $$ \int_X \sum_{k=0}^{\infty} |f_k(x)|=\sum_{k=0}^{\infty}\int_X |f_k| \,d \mu <\infty $$ by the Monotone convergence theorem. The inquality in (1) also implies that $F$ is integrable. To see why let $n\to\infty$ and integrate both sides.

Since $F_n\to F$, the DCT implies that $\int F_n\to \int F$ and hence $$ \int_X f d\mu = \sum_{n=0}^{\infty}\int_Xf_n\,d\mu $$

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For number two, note that \begin{align*} |f(x)|\leq\sum_{n=1}^{\infty}|f_{n}(x)|, \end{align*} so $\displaystyle\int|f(x)|dx\leq\int\displaystyle\sum_{n=1}^{\infty}|f_{n}(x)|dx=\sum_{n=1}^{\infty}\int|f_{n}(x)|dx<\infty$.

For number three, the dominated integrable function can be taken as $g(x):=\displaystyle\sum_{n=1}^{\infty}|f_{n}(x)|$.

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