Let $(X,\Sigma, u)$ be a measure space. Suppose $(f_n)$ is a sequence of measurable functions $f(x) = \sum_{n=1}^{\infty} f_n(x)$ converges $u$ almost everywhere and $|f(x)|\leq g$ for some $u$-summable function $g$.

I am trying to show that $\int_X f du = \sum_{n = 1}^{\infty} \int_X f_n du$.

I am pretty sure we must use the Lebesgue Dominated Convergence Theorem. We know that $\int_X |f| du \leq \int_X g du$ and $f$ is measurable as it is the $u$-almost everywhere pointwise limit of measurable functions. Hence $f$ is $u$-summable.

I want to consider the sequence $s_n= \sum_{k = 1}^{n} f_n $ since then $s_n\rightarrow f$. Now if we can show that $|s_n|<h$ for some $u$-summable function $h$, then the result will follow immediate from the dominated convergence theorem.

However I am struggling to show that $|s_n|$ can be dominated (of course things would be much easier if $f_n$ were positive). How can I show this?

  • $\begingroup$ In your second-last paragraph where you define $s_n$, do you mean to have $f_k$ in the sum? $\endgroup$
    – Glen_b
    Jun 25, 2017 at 23:39
  • $\begingroup$ Yes I do... thanks $\endgroup$
    – fosho
    Jun 26, 2017 at 6:21

2 Answers 2


You can't, it isn't true.

Take your favorite example of a sequence converging pointwise but not in $L^1$. For instance, let's take $X = [0,1]$ with Lebesgue measure; then the sequence $n 1_{[0,1/n]}$ will do. We want this to be $s_n$. Working backwards, we let $f_n = n 1_{[0,1/n]} - (n-1) 1_{[0, 1/(n-1)]}$ (with $f_1 = 1$). Then we have $\sum_{k=1}^\infty f_n(x) = 0$ almost everywhere, and $f=0$ is certainly dominated by a summable (i.e. integrable) function, with $\int f = 0$. On the other hand, $\int f_1 = 1$ and $\int f_k = 0$ for all $k \ge 2$, so $\sum_{k=1}^\infty \int f_k = 1$.

Generally speaking, having control on the limiting function $f$ is never enough in these situations (and the statement "$f$ is dominated by a summable function" is suspicious because it's the same as just saying "$f$ is summable"). You have to have some sort of uniform control over all the approximating functions $f_n$.

A typical condition under which this does hold is to have either $\sum_{k=1}^\infty \int |f_n| < \infty$, or $\int \sum_{k=1}^\infty |f_n| < \infty$ (you can use monotone convergence or Tonelli's theorem to see these are equivalent). In this case you can use either dominated convergence (with dominating function $\sum_{k=1}^\infty |f_n|$) or Fubini's theorem to conclude $\sum_{k=1}^\infty \int f_n = \int \sum_{k=1}^\infty f_n$.


This is wrong in general. Here is a counter-example. I will take $(X,\Sigma,\mu)=(\mathbb{R}, B_{\mathbb{R}},\lambda)$, and my sequeence of functions is composed of constant functions: $f_0(x)\equiv1$, and for $n\ge 1$, $f_n(x)\equiv \frac{1}{n+1}-\frac1n$.

The series $\sum_{n=0}^\infty f_n(x)$ converges every where to the constant function $f(x)\equiv0$, which is clearly integrable (one may take $g=f=0$). But clearly the equality $\sum_{n=0}^\infty \int_{\mathbb{R}}f_nd\lambda=0$ makes no sense.

Thus, there is a missing suplementary condition for the proposed equality to hold.


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