If $\{f_n\}$ is a Cauchy sequence in $(L^1,\|\cdot \|_{L^1})$ does it imply $f_n\to f$ for a certain $f\in L^1$?

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In the book "Real analysis" by Stein, the theorem 2.2 (Reisz-Fischer) says:

The vector space $L^1$ is complete in its metric.

To me, it means that if $(f_n)$ is a Cauchy sequence for $\|\cdot \|_{L_1}$, i.e. if $$\forall \varepsilon>0, \exists N\in\mathbb N: \forall n\in\mathbb N, n,m\geq N\implies \|f_n-f_m\|_{L^1}<\varepsilon$$ then there is a function $f\in L^1$ such that $$\forall \varepsilon>0, \exists N\in\mathbb N: \forall n\in\mathbb N, n\geq N\implies \|f_n-f\|_{L^1}<\varepsilon.$$

But does it imply that $f_n(x)\to f(x)$ a.e. ? I don't think is true but I can't find a counter exemple.