1
$\begingroup$

I am looking at the following problem:

Let $\{a_{n}\}$ be a sequence of nonnegative real numbers. Define the function $f$ on $E=[1,\infty)$ by setting $f(x) = a_{n}$ if $n \leq x < n+1$. Show that $\int_{E}f=\sum_{n=1}^{\infty}a_{n}$ using the Monotone Convergence Theorem.

Now, if we knew that $f$ were measurable, we could use additivity over domains of integration, and the problem would be nice.

However, we cannot, and I can't for the life of me figure out how to use the Monotone Convergence theorem to do this problem. The $\{a_{n}\}$ aren't increasing, so what about this calls for the Monotone Convergence Theorem?? How do I go about proving this?

$\endgroup$
  • 2
    $\begingroup$ Let $f_n=a_n\chi_{[n,n+1)}$. Then $f_n$ is certainly measurable. And $f=\sum f_n$ so $f$ is measurable. Apply MCT to the partial sums of $\sum f_n$. $\endgroup$ – David C. Ullrich Nov 11 '15 at 15:23
2
$\begingroup$

The key point, that is never stressed enough, is what $\sum_{n=1}^\infty a_n$ means. It means, by definition, $$ \lim_{N\to\infty}\sum_{n=1}^Na_n. $$ So this is the limit of an increasing sequence of real numbers. So, $$ \sum_{n=1}^\infty a_n=\lim_{N\to\infty}\sum_{n=1}^Na_n=\lim_{N\to\infty}\int_E f_n=\int_E f, $$ where $f_n=\sum_{n=1}^N a_n\,1_{(n,n+1]}$ gives an increasing sequence of functionsand the last equality is given by Monotone Convergence.

$\endgroup$

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.