$\sum_n(-1)^n\int_Xf_n \, d\mu=\int_X\sum_n(-1)^nf_n \, d\mu$

Question

Let $(X,\mathcal{F},\mu)$ be measure space, $(f_n)_n$ be a non-increasing sequence of functions in $L^1(\mu)$, converging $\mu$-a.e to $0.$

Prove that $$\sum_n(-1)^n\int_Xf_n \, d\mu=\int_X\sum_n(-1)^nf_n \, d\mu.$$

To prove it, it's evident that we should use the dominated convergence theorem on $\sum_{k=1}^n(-1)^kf_k:$

$$\sum_{k=1}^n(-1)^k\int_Xf_k \, d\mu=\int_X\sum_{k=1}^n(-1)^kf_k \, d\mu$$ Since $(f_n)_n$ is a non-increasing sequence converging $\mu$-a.e to $0$ (which means $f_n$ are non-negative $\mu$-a.e), this shows that $(\sum_{k=1}^n(-1)^kf_k)_n$ converges $\mu$-a.e (to $\sum_n(-1)^nf_n$), it remains to prove that $\sum_{k=1}^n(-1)^kf_k$ is dominated by a function $\phi \in L^1.$

Do you know how to have a verification?

Answer 1

Since $0\leq f_{n+1}\leq f_n$, we have that $S=\sum_n(-1)^{n+1}f_n$ converges $\mu$-al,ost surele. Also, by well known properties of alternating series with nondecreasing terms we have that

$$ |S-S_n|=(-1)^n(S-S_n)\leq f_{n+1}\leq\ldots\leq f_1 $$ By dominated convergence

1. $\|S-S_n\|_1\xrightarrow{n\rightarrow\infty}0$.
2. $S\in L_1$ since $|S|\leq |S-S_n|+f_{n+1}$.
3. $\int S\,d\mu = \sum_n(-1)^{n+1}\int f_n\,d\mu$