1
$\begingroup$

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!!!

$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Catherine Drysdale May 12 '17 at 15:11
  • $\begingroup$ Upward arrow just means that the sequence is increasing and converging. $\endgroup$ – Karthik P N May 12 '17 at 15:11
  • $\begingroup$ but what does := mean? And why do we need to write g in terms of h? $\endgroup$ – Catherine Drysdale May 12 '17 at 15:12
  • $\begingroup$ Sorry for messing up with the notations. I've edited my answer. Is it understandable now? $\endgroup$ – Karthik P N May 12 '17 at 15:13
  • $\begingroup$ yes, I think so, but it E just a constant to make up for the partition being finite now? $\endgroup$ – Catherine Drysdale May 12 '17 at 15:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.