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Let $f_n: \mathbb{R} \to \mathbb{R}$ be measurable for $n = 1,2,\dots$ Let $a_n = \int_{\mathbb{R}} |f_n|$ for $n = 1,2,\dots$ and assume that $\sum_{n = 1}^{\infty} a_n < \infty$. Prove that $\sum_{n = 1}^{\infty} f_n$ converges almost everywhere.

This essentially becomes $\sum_{n = 1}^{\infty} \int_{\mathbb{R}} |f_n| < \infty$. The question is, can one interchange the integral and the sum? We've talked so much about interchanging limits and integrals with the convergence theorems, but what about for a sum? What would (or would not) allow me to do that here? Is that the right approach? Thanks!

Nolan P
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  • Check out the first answer here (or the second one that uses the MCT): https://math.stackexchange.com/questions/83721/when-can-a-sum-and-integral-be-interchanged. After interchanging the integral and sum you still would need to explain why that implies convergence almost everywhere of the series. – folouer of kaklas Apr 25 '21 at 19:56
  • Well I think it's obvious at that point. You have $\int \sum_{n = 1}^{\infty} |f_n|| < \infty$. One could just take the derivative of both sides to remove the integral, and you'd have the result. @folouerofkaklas – Nolan P Apr 25 '21 at 21:45
  • Maybe I'm misunderstanding you, but you can't take the derivative of that. You could consider the set where the series doesn't converge and show it has measure zero. – folouer of kaklas Apr 25 '21 at 23:19
  • why couldn't you? It's an integral, so it would cancel out? – Nolan P Apr 26 '21 at 12:14
  • The limits of integration are fixed. – folouer of kaklas Apr 26 '21 at 12:45
  • @folouerofkaklas oh you're right. Hmm, well how would I show where the series doesn't convergence has measure 0? – Nolan P Apr 26 '21 at 19:18

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By using the triangle inequality and monotonicity of integrals, and then MCT (see the addendum below) to exchange summation and integral we obtain $$\int\bigg|\sum_{n=1}^{\infty}f_n\bigg|dx\leq \int\sum_{n=1}^{\infty}|f_n|dx\underbrace{=}_{\textrm{MCT}}\sum_{n=1}^{\infty}\int|f_n|dx<\infty$$ Since this implies $|f|=|\sum_{n \in\ \mathbb{N}}f_m| \in \mathcal{L}^1(dx)$, by the properties of functions in $\mathcal{L}^1$ then $$\implies f=\sum_{n=1}^{\infty}f_n \in \mathcal{L}^1(dx)$$ which means that $f$ is a.e. $\mathbb{R}$-valued.

As an addendum, it is straightforward to see how MCT leads to the result about sums. Set $s_M=\sum_{n=1}^M|f_n|$, (partial sums as a sequence in $\mathbb{N}$) you have $$\int \sup_{M \in \mathbb{N}}s_M d\mu = \sup_{M \in \mathbb{N}} \int s_M d\mu = \sup_{M \in \mathbb{N}} \int \sum_{n=1}^M|f_n|d\mu \underbrace{=}_{\textrm{linearity}} \sup_{M \in \mathbb{N}}\sum_{n=1}^M\int |f_n|dx$$ and $\sum_{n \in \mathbb{N}}|f_n|\in \mathcal{L}^1(dx)$ because it has been assumed that $$\sup_{M \in \mathbb{N}}\sum^M_{n=1}\int|f_n|d\mu=\sum_{n=1}^\infty\int|f_n|d\mu<\infty$$

Snoop
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