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In analysis courses we are taught that we must require two conditions to be true for a series of functions and an integral to be interchanged, ie if $f_n(x)$ is some series of functions with

(A) $\sum_{n = 1}^{\infty} |f_n(x)|$ converges

(B) $\sum_{n = 1}^{\infty} \int_I |f_n(x)|$ converges

then we can exchange the integral and the sum:

$$ \int_I \sum_{n = 1}^{\infty}f_n(x) \; = \; \sum_{n = 1}^{\infty} \int_I f_n(x) $$

Why is condition (B) necessary? Specifically, can someone give a counter example to the theorem with condition (B) removed, ie give a series of functions $(f_n(x))$ where $ \int_I \sum_{n = 1}^{\infty}f_n(x) $ and $ \sum_{n = 1}^{\infty} \int_I f_n(x) $ converge to different values because condition (B) is not satisfied?

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  • $\begingroup$ (B) is not necessary, but it does imply (A). $\endgroup$
    – fwd
    Commented Nov 3, 2021 at 15:55

1 Answer 1

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The condition (B) implies that the function $$ g(x) := \sum_{n=1}^{\infty} |f_n(x)| $$ is integrable, since the monotone convergence theorem implies that $$ \int\sum_{n=1}^{\infty} |f_n(x)| \ dx = \sum_{n=1}^{\infty}\int|f_n(x)| \ dx. $$

If the RHS is finite, it means $g(x)$ is integrable and hence finite a.e.

If you allowed integrals in the extended sense, then (B) would be unnecessary. Otherwise, whenever the RHS is finite, it must equal the LHS.

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