I am expected to use the Monotone convergence theorem for non-negative functions in order to solve this question. I.e. LEt $(X, \Sigma, \mu)$ me a measure space. Suppose $fn:X \rightarrow \bar{\mathbb{R}}$ is a sequence of non-decreasing, non-negative measurable functions. Then $\int \lim_{n\rightarrow \infty}f_nd\mu = \lim_{n\rightarrow \infty} \int f_n d\mu$.

Assume $g_n$ is a sequence of non-negative measurable functions satifying$\int g_n d\mu < \frac{1}{n^2}$ for each $n\geq 1$. Prove that $\Sigma_{n=1}^{\infty}g_n(x) \leq + \infty$.

My thinking was that $\Sigma_{n=0}^{\infty} \int g_n < \Sigma_{n=0}^{\infty} \frac{1}{n^2}$, but because the integral amounts to summation anway we could swap the sum and the integral around... This seems dodgy and it doesn't use the monotone convergence theorem though! Help!!!


1 Answer 1


Here, it is not the question of integral amounting to summation. Notice that

\begin{equation} \sum\limits_{n=1}^{\infty}g_{n}(x)=\lim\limits_{k\to \infty}\sum\limits_{n=1}^{k}g_{n}(x). \end{equation} The key idea is that $\sum\limits_{n=1}^{k}g_{n}(x)\uparrow \sum\limits_{n=1}^{\infty}g_{n}(x)$ as $k\to \infty$, which is why the limit on the right hand side of the above equation I have written always exists (could perhaps be $+\infty$) since it is a limit of a sequence of nondecreasing positive numbers. Now, monotone convergence theorem tells us that

\begin{align} \int \sum\limits_{n=1}^{\infty}g_{n}(x)\,d\mu(x)&=\int \lim\limits_{k\to \infty}\sum\limits_{n=1}^{k}g_{n}(x)\,d\mu(x)\stackrel{MCT}{=}\lim\limits_{k\to \infty}\int\sum\limits_{n=1}^{k}g_{n}(x)\,d\mu(x)\\&=\lim\limits_{k\to \infty}\sum\limits_{n=1}^{k}\int g_{n}(x)\,d\mu(x)\leq \sum\limits_{n=1}^{\infty}\frac{1}{n^{2}}<\infty. \end{align}

So, what we have shown is that $\int \sum\limits_{n=1}^{\infty}g_{n}(x)\,d\mu(x)<\infty$. Now, using the fact that for a nonnegative random variable $X$, $\int X\,d\mu<\infty$ implies $X<\infty$ a.e., we arrive at the conclusion that $\sum\limits_{n=1}^{\infty}g_{n}<\infty$ a.e.

  • $\begingroup$ I am sorry, I don't really undestand some of the notation - for instance, what does the upwards arrow mean and what does =: mean? x $\endgroup$ May 12, 2017 at 15:11
  • $\begingroup$ Upward arrow just means that the sequence is increasing and converging. $\endgroup$ May 12, 2017 at 15:11
  • $\begingroup$ but what does := mean? And why do we need to write g in terms of h? $\endgroup$ May 12, 2017 at 15:12
  • $\begingroup$ Sorry for messing up with the notations. I've edited my answer. Is it understandable now? $\endgroup$ May 12, 2017 at 15:13
  • $\begingroup$ yes, I think so, but it E just a constant to make up for the partition being finite now? $\endgroup$ May 12, 2017 at 15:14

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