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I am trying to prove the following

Theorem: Let $\{A_1, A_2, \cdots \}$ be a countable disjoint collection of sets in $\mathbb R$ and let $S = \bigcup_{i=1}^\infty A_i$. Let $f$ be defined on $S$.

(a) If $f\in L(S)$, then $f\in L(A_i)$ for each $i$ and $$\int_S f = \sum_{i=1}^\infty \int_{A_i} f.$$ (b) If $f\in L(A_i)$ for each $i$ and if the series in (a) converges, then $f\in L(S)$ and the equation in (a) holds.

(a) is easy, but I am not able to prove part (b) as I don't know what result previously proved can be used because $f$ is not a sequence.

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    $\begingroup$ (b) is not true. (It is true if $f \geq 0$.) It is easy to construct a function $f$ such that $\int_n^{n+1}f(x)dx=0$ for all $n$ but $f$ is not integrable on the union of the intervals $(n, n+1)$. $\endgroup$ Jun 29, 2020 at 8:03
  • $\begingroup$ @Thanks for telling. Can you kindly outline a proof if f$\geq$0 ? It would be really helpful as I am struck on it from many days. $\endgroup$
    – Avenger
    Jun 29, 2020 at 8:17

1 Answer 1

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Proof of (b) when $f \geq 0$. [As already pointed out (b) is false in general].

Let $f_n= \sum\limits_{i=1}^{n} f\chi_{A_i}$. Then $(f_n)$ is a sequence of non-negative measurable functions increasing to $f$. By Monotone Convergence Theorem we have $\int_S f= \lim \int_A f_n=\lim \sum\limits_{i=1}^{n} \int_{A_i} f$. Hence $\int_S f <\infty$ and $\int_S f =\sum\limits_{i=1}^{\infty} \int_{A_i} f$.

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  • $\begingroup$ thanks a lot for your help. $\endgroup$
    – Avenger
    Jun 29, 2020 at 8:51

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