Prove that for all integrable functions $g_n, g$, we have the implication $\|g_n-g\|_1\to 0\Rightarrow \|g_n\|_1\to \|g\|_1$ as $n\to \infty$. Is the converse true?
It seems like $|g_n-g|_1 \to 0$ implies that $|\int_X g_n dm - \int_x g dm| \to 0$, since the latter is bounded by $\int_X|g_n-g| dm$. Is it okay to just take the limit and call it good? I feel like I am missing some details here. Also, what about the converse?
On
$(\Rightarrow)$ By the reverse triangle inequality $$ | \|g_n\| -\|g\| | \leq \| g_n -g \| $$ Since $\| g_n -g \| \to 0$ then clearly $\|g_n\| \to \|g\|$
The converse is also true as long $g, g_n \in L ^1$ and $g_n \to g$ a.e. In that case you can use the next argument:
$(\Leftarrow)$ First note that since $|g_n -g| \leq |g_n| + |g|$ then $|g_n -g| \in L^1$. Also $ |g_n| + |g| \to 2|g|$ a.e and by hypothesis $\int |g_n| + |g| dm \to \int 2|g| dm$, moreover $|g-g_n| \to 0 $ a.e, thus by a generalized version of the dominated convergence theorem we conclude that $$ \|g_n - g \| \to 0 $$
It looks like your reasoning is fine, except that you may want to add some more detail as to how $$\left|\, \int_{X} g_n \, dm - \int_{X}g\, dm \,\right|$$ is related to the (absolute value of) the difference of the norms. Specifically, I’d write something like $$\left|\, ||g_n||_{1} - ||g||_1 \,\right| = \left|\, \int_{X} |g_n| \, dm - \int_{X} |g| \, dm \,\right| \leq \int_{X} \left| \,|g_n| - |g| \,\right| \, dm \leq \int_{X} |g_n - g| \, dm.$$
The converse is false. As a counterexample, let $X = [0, 1]$ with Lebesgue measure. Let $g_n = 2 \, \chi_{[0,1/2]}$ (two times the characteristic function of $[0, 1/2]$) if $n$ is even, and let $g_n = 2\chi_{[1/2, 1]}$ if $n$ is odd. Finally, let $g(x) = 1$ for all $x \in [0,1]$.