$\{g_n\}_1^\infty \subset L^1(X,\mathcal F,\mu)$ and $\sum_1 ^\infty|g_n| < \infty$ prove $G=\sum_1 ^\infty g_n \in L^1$

Question

I started this way:

$\int |\sum_1^\infty g_n| d\mu \le \int \sum_1^\infty |g_n| d\mu $ now I want to show that the last integral is finite.

$\int \sum_1^\infty |g_n|d\mu = \int\lim_{k\rightarrow \infty} \sum_1^k|g_n| =_{M.C.T} \lim_{k\rightarrow \infty}\int\sum_1^k|g_n|$

But here I'm stuck cause the last limit might be $\infty$ .

Moreover I'm not sure whether the M.C.T usage is correct cause $g_n$ are $L^1$ function therefore I can't really say anything about specific $x's$ value.

Answer 1

Not true. Consider $g_n(x)=\frac{1}{n^2x}$ on the measure space $[1,\infty)$ with the Lebesgue measure. For every $x\geq 1$: $$\sum_{n=1}^{\infty}|g_n(x)|=\sum_{n=1}^{\infty}g_n(x)=\frac{\pi^2}{6x}<\infty$$ but $\frac{\pi^2}{6x}\notin L_1$.

Answer 2

Let $X = \mathbb{R}$, $\mathcal{F} = \mathscr{B}(\mathbb{R})$ and let $\mu = m$

Pick $g_n(x) = 1_{[n-1, n)}$, so $g_n \in L^1$, since, $$\int\limits_{\mathbb{R}}g_n \mathrm{d}\mu = 1 < \infty$$ Moreover, $\sum\limits_{n=1}^{\infty}|g_n| = 1 < \infty$, but $1_{[0, \infty)} = \sum\limits_{n=1}^{\infty} g_n$ is not (finitely) integrable