# Convergence of Lebesgue integrable functions in an arbitrary measure.

I'm a bit stuck on this problem, and I was hoping someone could point me in the right direction.

Suppose $f, f_1, f_2,\ldots \in L^{1}(\Omega,A,\mu)$ , and further suppose that $\lim_{n \to \infty} \int_{\Omega} |f-f_n| \, d\mu = 0$. Show that $f_n \rightarrow f$ in measure $\mu$.

In case you aren't sure, $L^1(\Omega,A,\mu)$ is the complex Lebesgue integrable functions on $\Omega$ with measure $\mu$.

I believe I have to use the Dominated convergence theorem to get this result, and they usually do some trick like taking a new function $g$ that relates to $f$ and $f_n$ in some way, but I'm not seeing it. Any advice?

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We wish to show that for any $\epsilon > 0,$ there is some $N$ such that for all $n \geq N, \mu(\{x : |f_n(x) - f(x)| > \epsilon\}) < \epsilon.$ (This is one of several equivalent formulations of convergence in measure.)
If this were not the case, then there'd be some $\epsilon > 0$ so that for every $N$ there is an $n \geq N$ that doesn't satisfy the above condition. So, pick $N$ large enough so that for all $n \geq N, \int_\Omega |f-f_n| \ d\mu < \epsilon^2.$ Then, for this $N$ we have, by our assumption, some $n_0$ with $\mu(L_{n_0}) \geq \epsilon$ where $L_{n_0} = \{x: |f_{n_0}(x) - f(x)| > \epsilon\}.$
But then, we'd have that $$\epsilon^2 > \int_\Omega |f_{n_0} - f| \ d\mu \geq \int_{L_{n_0}} |f_{n_0} - f| \ d\mu > \epsilon\mu(L_{n_0}) \geq \epsilon^2,$$ which is a contradiction. Hence, we must have convergence in measure.