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Prove.

If $f_k$, $k=1, 2, \cdots,$ are nonnegative, measurable, and defined on $E\subset\mathbb{R}^n$, then $$\int_E{\left(\sum_{k=1}^{\infty}f_k\right)}=\sum_{k=1}^{\infty}\int_E{f_k}$$

Proof.

  1. Let $F_N$ be a function defined by $\displaystyle F_N=\sum_{k=1}^{N}f_k$
  2. Then, $F_N$ are nonnegative and measurable, and increase to $\displaystyle\sum_{k=1}^{\infty}f_k$
  3. Hence $$\int_E{\left(\sum_{k=1}^{\infty}f_k\right)}=\lim_{N\to\infty}\int_EF_N=\lim_{N\to\infty}\sum_{k=1}^{N}\int_Ef_k=\sum_{k=1}^{\infty}\int_Ef_k$$

I think the fact that $\displaystyle F_N\nearrow{}F_\infty$ is required, in order to be $$\int_E{\left(\sum_{k=1}^{\infty}f_k\right)}=\int_E{F_\infty}=\int_E\lim_{N\to\infty}F_N=\lim_{N\to\infty}\int_EF_N$$ since Monotone Convergence Theorem gives a sufficient condition for interchanging the operations of integration and passage to the limit.

Where and how can I get the fact that $\displaystyle F_N\nearrow{}F_\infty$.

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  • $\begingroup$ Each $f_k$ is nonnegative so $F_N$ increases with $N$. $\endgroup$
    – yurnero
    May 23, 2016 at 1:43
  • $\begingroup$ @yurnero I know that $F_1{\le}F_2{\le}\cdots{\le}F_N\le\cdots$. However, I do not agree with that $F_k$ can be converged because it can increase endlessly. If so, we can't say that it converges, can we? $\endgroup$
    – Danny_Kim
    May 23, 2016 at 1:48
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    $\begingroup$ For each $x$, either $\sup_n f_n(x)<\infty$ or $\sup_n f_n(x)=\infty$. In either case, the limit is well-defined. $\endgroup$
    – Math1000
    May 23, 2016 at 2:23
  • $\begingroup$ @Math1000 Ah, you mean it's fine not to care about the case when $f_k$ increases ceaselessly, right??? The bell is rung in my brain. Thank you. $\endgroup$
    – Danny_Kim
    May 23, 2016 at 2:26
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    $\begingroup$ it is obvious that when we impose non-negativity, there is no conditional convergence anymore, and everything is absolutely convergent in $\mathbb{R}^+ \cup \{+\infty\}$, whatever the order of summation $\endgroup$
    – reuns
    May 23, 2016 at 2:27

1 Answer 1

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Background information: $L^+$ is the space of all measurable functions from $X$ to $[0,\infty]$.

Theorem 2.13 (Real Analysis, Folland) - Let $\phi$ and $\psi$ be simple functions in L^+$

b.) $\int (\phi + \psi) = \int \phi + \int \psi$

If $\{f_n\}$ is a finite or infinite sequence in $L^{+}$ and $f = \sum_{n}f_n$, then $\int f = \sum_{n}\int f_n$

Proof - Let $f,g\in L^{+}$, and suppose $\phi_n\rightarrow f, \psi_n\rightarrow g$ monotonically. Applying theorem 2.13b we get $$\int f + \int g = \lim_{n\rightarrow \infty}\int \phi_n + \lim_{n\rightarrow \infty}\int \psi_n = \lim_{n\rightarrow \infty}\int \phi_n + \int \psi_n$$ $$= \lim_{n\rightarrow \infty}\int (\phi_n + \psi_n) = \int (f+g)$$ Hence by induction, $\int\sum_{1}^{n}f_n = \sum_{1}^{n}\int f_n$ for any finite $n$. Let $n\rightarrow \infty$, by the monotone convergence theorem $$\int \sum_{1}^{\infty}f_n = \sum_{1}^{\infty}\int f_n$$

I am not 100% sure this is entirely relevant to your question. I will delete it if that happens to be the case.

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