# Counter example to exchanging summation and integral

In analysis courses we are taught that we must require two conditions to be true for a series of functions and an integral to be interchanged, ie if $$f_n(x)$$ is some series of functions with

(A) $$\sum_{n = 1}^{\infty} |f_n(x)|$$ converges

(B) $$\sum_{n = 1}^{\infty} \int_I |f_n(x)|$$ converges

then we can exchange the integral and the sum:

$$\int_I \sum_{n = 1}^{\infty}f_n(x) \; = \; \sum_{n = 1}^{\infty} \int_I f_n(x)$$

Why is condition (B) necessary? Specifically, can someone give a counter example to the theorem with condition (B) removed, ie give a series of functions $$(f_n(x))$$ where $$\int_I \sum_{n = 1}^{\infty}f_n(x)$$ and $$\sum_{n = 1}^{\infty} \int_I f_n(x)$$ converge to different values because condition (B) is not satisfied?

• (B) is not necessary, but it does imply (A).
– fwd
Commented Nov 3, 2021 at 15:55

The condition (B) implies that the function $$g(x) := \sum_{n=1}^{\infty} |f_n(x)|$$ is integrable, since the monotone convergence theorem implies that $$\int\sum_{n=1}^{\infty} |f_n(x)| \ dx = \sum_{n=1}^{\infty}\int|f_n(x)| \ dx.$$
If the RHS is finite, it means $$g(x)$$ is integrable and hence finite a.e.