Assuming $\sum_{n = 1}^\infty \int |f_n| < \infty$, properties that follow for integral

Question

How do I see that if $\sum_{n = 1}^\infty \int |f_n| < \infty$, then

1. $\sum_{n = 1}^\infty f(x)$ converges absolutely almost everywhere,
2. is integrable, and
3. its integral is equal to $\sum_{n = 1}^\infty \int f_n$?

Thanks in advance!

Answer 1

We have:

$$\int \sum_{n=1}^{\infty} |f_n| = \sum_{n=1}^{\infty} \int |f_n| < \infty$$

Hence, $\sum_{n=1}^{\infty} |f_n|$ is finite a.e. (i.e. $\sum f_n$ converges absolutely a.e.)

Now:

$$\int \left| \sum_{n=1}^{\infty} f_n \right| \le \int \sum_{n=1}^{\infty} |f_n| < \infty$$

Finally, note that the sequence $(\sum_{k=1}^n f_k )_n$ converges a.e. to $\sum_{n=1}^{\infty} f_n$, and is dominated by $\sum_{n=1}^{\infty} |f_n|$ (which is in $L^1$). Apply DCT to get $3.$

Answer 2

By MCT, $$ \int\sum_{n=1}^\infty|f_n| = \sum_{n=1}^\infty \int|f_n| <+\infty. $$ Hence $\sum_n |f_n(x)|$ converges a.e., and so does $\sum_n f_n(x)$. The other statements follow from DCT.

Note that technically you need to state that $f_n$ are measurable.