Let $(X,\Sigma,\mu)$ be a finite measure space. $f_n,f:X \to\mathbb{R}$ such that there is $M\in\mathbb{R}$ where $|f_n(x)|,|f(x)|\le M$
prove that if $f_n\to f$ in measure than $f_n\to f$ in $L^1(\mu)$
My attempt:
Let $A=\{x\in X|f_n(x)-f(x)|\ge \epsilon \}$ than $\lim\mu(A)=0$ and also: $\int_X|f_n(x)-f(x)|d\mu \le \int_A|f_n(x)-f(x)|d\mu \le \int_A|f_n(x)|+|f(x)|d\mu \le\int_A2Md\mu=2M\mu (A)$ now if we take the limit we get the result.
Is this correct? seems too simple too me, but I hope I'm right.
Notice that you can write $$\int_X |f_n - f| d \mu = \int_{A_n} |f_n - f| d \mu + \int_{X \setminus A_n}|f_n - f| d \mu$$ for $A_n = \{x \in X: |f_n(x) - f(x)| \geq \varepsilon\}$ so that $\mu(A_n) \to 0$ as $n \to \infty$.
Then, for $x \in X \setminus A_n$, you have $|f_n(x) - f(x)| < \varepsilon$ and so you can bound \begin{align*} \int_X |f_n - f| d \mu &\leq \int_{A_n} 2M d \mu + \int_{X \setminus A_n} \varepsilon d \mu \\& = 2M \mu(A_n) + \mu(X \setminus A_n) \varepsilon \\& \leq 2M \mu(A_n) + \mu(X) \varepsilon \end{align*} Since $\varepsilon$ is arbitrary, sending $\varepsilon \to 0$ gives $$\int_X |f_n - f| d \mu \leq 2M \mu(A_n) \to 0$$ as $n \to \infty$.