Let $\{f_n\}$ be a sequence of Lebesgue integrable functions that converges to $f$ in $L^1$. Then, I have to show that $\{f_n\}$ converges in measure.
Here is my approach:
So Claim: $\mu(x:|f_n(x)-f(x)|\geq\epsilon) \to 0$
for any $\epsilon >0$, by Chebyshev's Inequality, we have
Case 1: Consider $\mu(x:|f_n(x)-f(x)|> \epsilon)<\frac{1}{\epsilon}\int{|f_n(x)-f(x)|d\mu} \to0$ (as $f_n \to f$ in $L^1)$
Case 2: Consider $\mu(x:|f_n(x)-f(x)|=\epsilon)=\mu(x:-\epsilon +f(x)\leq f_n(x)\leq\epsilon+f(x)) \to 0(???)$ for any arbitrary $\epsilon$.
I am not sure about Case 2.
Thanks for any help!!
There's not really any reason to have case 2, because Chebyshev's inequality also works with $\le$ replacing $<$. If you're dead set on using it, then you haven't got the setup quite right (your first equality is wrong). The relevant estimate would be that
$$\mu\{x : g(x) = \lambda\} = \frac 1 {\lambda} \int_{\{x : g(x) = \lambda\}} g\, d\mu \le \frac{1}{\lambda} \int g \, d\mu$$ for positive $g$.
Alternatively, notice that
$$\mu\{|f_n - f| \ge \epsilon\} \le \mu\left\{|f_n - f| > \frac{\epsilon}{2}\right\}$$
and apply Case $1$.