Prove that convergence of the infinite series of integral of absolue values of a sequence of functions implies convergence

Let $S_{N}=\sum_{n=1}^{N}f_{n}$. To prove that $\sum_{n=1}^{\infty}f_{n}$ it is enough to show that $\{S_{N}\}$ is Cauchy sequence in $L^1$. 

Let $\epsilon>0$. Then there exists $M$ such that 
\[\sum_{n=M+1}^{N}\int|f_{n}|<\epsilon\]

For $N>M$, then
$$\|S_{N}-S_{M}\|_{L^{1}}=\int|S_{N}-S_{M}|=\int|\sum_{n=M+1}^{N}f_{n}|\leq\int\sum_{n=M+1}^{N}|f_{n}|=\sum_{n=M+1}^{N}\int|f_{n}|<\epsilon.$$

Hence $\{S_{N}\}$ is convergent in $L^{1}$, say, $\|S_{N}-f\|_{L^{1}}\rightarrow 0$ for some $f\in L^{1}$. This means that $\sum_{n=1}^{\infty}f_{n}$ converges to some function $f\in L^1$. 

Note also that $\sum_{n=1}^{\infty}\int|f_{n}|=\int\sum_{n=1}^{\infty}|f_{n}|$ and so $\int\sum_{n=1}^{\infty}|f_{n}|<\infty$, this entails $\sum_{n=1}^{\infty}|f_{n}(x)|<\infty$ a.e. and hence $\sum_{n=1}^{\infty}f_{n}(x)$ exists a.e.

Also 
\[0=\lim _{N\rightarrow \infty} \|S_{N}-f\|_{L^{1}}=\lim _{N\rightarrow \infty} \int |\sum_{n=1}^{N}f_{n} -\sum_{n=1}^{\infty}f_{n}(x)|\geq \lim _{N\rightarrow \infty}| \int \sum_{n=1}^{N}f_{n} - \int \sum_{n=1}^{\infty}f_{n}(x)|\]
\[=\lim _{N\rightarrow \infty}| \sum_{n=1}^{N} \int f_{n} - \int \sum_{n=1}^{\infty}f_{n}(x)|\]
\[=|\lim _{N\rightarrow \infty} \sum_{n=1}^{N} \int f_{n} - \int \sum_{n=1}^{\infty}f_{n}(x)|\]
\[=| \sum_{n=1}^{\infty} \int f_{n} - \int \sum_{n=1}^{\infty}f_{n}(x)|=0.\]
This means 
\[\sum_{n=1}^{\infty} \int f_{n} = \int \sum_{n=1}^{\infty}f_{n}(x)\]