Let $P(N)$ be the power set of $N$ and $u$ be the counting measure on $N$.
(a) Prove or disprove the measure space $(N, P(N), u)$ is complete?
(b) Given function $g: N\rightarrow R$. Show that $g$ is integrable if and only if $\sum_{m=1}^{\infty} g(m)$ is absolutely convergent.
My attempt: For (a), I was able to show that due to definition of $u$, the only set $E\in N$ with $u(E) = 0$ is the empty set. And the empty set belongs to $P(N)$, thus the measure space $(N, P(N), u)$ is complete.
For (b), by using two results: (1) "For nonnegative, measurable function $f$ in the measure space $(N, P(N), u)$, we have: $\int_{N} f du = \sum_{n=1}^{\infty} f(n)$, and (2) "f is integrable iff $|f|$ is integrable," we apply it with the function $g$ (note that $g$ is always $u$-measurable in the given measure space), then it's sufficient to show that $|g|$ is integrable iff $\sum_{m=1}^{\infty} |g(n)|$ is absolutely convergent. By using result (1), since $\int_{N} g du = \sum_{m=1}^{\infty} |g(n)|$, we're done.
Can someone please help review if my solutions above is correct, especially for part (b). The result (1) is quite strong, since it doesn't require $f$ to be integrable at all (in fact, if $f$ is not integrable, then the series $\sum_{n=1}^{\infty} f(n)$ is divergent, but the equality still holds).