# Lebesgue integral and sums

I'm not particularly well read on the Lebesgue integral, but I have heard that it permits a much wider class of functions, and in particular we can interchange integrals and limits more easily. However, I was wondering, are there any similar benefits with the Lebesgue integral when it comes to sums and integrals. For example, does it make it "easier" to interchange an infinite sum and integral sign?

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Yes, it does.

Case 1: If $\{f_n\}$ are nonnegative measurable functions, then: $$\int_X \sum_{n=1}^\infty f_n \, d\mu = \sum_{n=1}^\infty \int_X f_n \, d\mu$$

In other words, you can always interchange an infinite sum and the integral sign when dealing with nonnegative functions. This is an immediate result of the monotone convergence theorem.

Case 2: If $\{f_n\}$ are complex measurable functions and: $$\sum_{n=1}^\infty \int |f_n| \, d\mu < \infty$$

Then the series $\sum_{n=1}^\infty f_n(x)$ converges for almost all $x$, and we have: $$\int_X \sum_{n=1}^\infty f_n \, d\mu = \sum_{n=1}^\infty \int_X f_n \, d\mu$$

This is a result of the dominated convergence theorem.

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So in case 2 we could in fact have, supposing $f_n$ are purely real, that $f_n$ could take on both positive and negative values but still the integral and sum could be interchanged if the condition you state is met? – poirot Dec 12 '12 at 22:18
@pbs Indeed. The result still holds if $\{f_n\}$ are real functions. – Ayman Hourieh Dec 12 '12 at 22:20
Am I right in thinking that the Lebesgue integral with the Lebesgue measure is just a Riemann integral? And if so then we can use the above theorems on these Riemann integrals? – poirot Dec 13 '12 at 8:25
@pbs If a bounded function is Riemann integrable, it's also Lebesgue integrable and the integrals coincide. So everything one learns in elementary Calculus is still useful in the context of the Lebesgue integral, and the convergence theorems above can sometimes be applied to the Riemann integral. – Ayman Hourieh Dec 13 '12 at 9:27
@pbs However, the Lebesgue integral is more flexible and works for a larger class of functions (even with the Lebesgue measure). Check out this example in which the monotone convergence theorem fails for the Riemann integral. – Ayman Hourieh Dec 13 '12 at 9:27

Yes. For example, if you have a sequence of nonnegative measurable functions $f_1, f_2, \ldots$, then it is always true that $$\sum_{j=1}^\infty \int_{\Omega} f_j\, d\mu=\int_{\Omega} \sum_{j=1}^\infty f_j\, d\mu.$$ This is one of the consequences of the monotone convergence theorem and it is easy to prove: the sequence of the partial sums $\sum_1^n f_j(x)$ is monotone with respect to $n$ at almost all $x\in \Omega$.

You do not have an analogous result in Riemann theory.

EDIT As anonymous points out in comments, you do have an analogous result for the Riemann integral, that is

If a sequence $f_1, f_2 \ldots$ of nonnegative Riemann integrable functions is such that the series
$$\tag{!}\sum_{j=1}^\infty f_n(x)\ \text{is Riemann integrable},$$ then you can interchange sum and integral.

The main problem with this theorem is the assumption marked with (!): in general, that series (even if convergent) needs not be Riemann integrable.

This kind of phenomenon can be compared with the non-completeness of the rational system. Namely, a series of nonnegative rational numbers needs not converge in $\mathbb{Q}\cup\{+\infty\}$: take for example $\sum n^{-2}$.

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You don't? I'm fairly certain that if $\{f_n\}_{n=1}^{\infty}$ is a sequence of nonnegative Riemann integrable functions on some interval $[a,b]$ which sum pointwise to a Riemann integrable function, then the sum and (Riemann) integral may be interchanged. – anonymous Dec 12 '12 at 22:42
@anonymous: Of course, but you need to know a priori that the sum is Riemann integrable. (And even in that case I am not sure that the proof is so straightforward - is it?). What I wanted to say with that final remark is that here you basically need nothing. Seeing your comment I'm afraid that remark of mine is a bit misleading. – Giuseppe Negro Dec 12 '12 at 22:52
Oh, yes, sorry about the misunderstanding. I just wanted to point out that it's not the results that aren't true, just that the proofs are more difficult and that pointwise limits preserve measurability. – anonymous Dec 12 '12 at 22:58