Corollaries of Chebyshev inequality in measure theory

Question

I would like to share proofs and hopefully get some feedback.

1. Show that if $f$ is integrable, then the set $A:=\{x\mid |f(x)|>0\}$ is a countable union of sets of finite measure.

2. Show that the set $B:=\{x\mid f(x)=\infty\}$ is null.

My attempt:

1. We have $A=\bigcup^{\infty}_{k=1}A_k$, where $A_k=\{x\mid |f(x)|\geq 1/k\}$. Each $A_k$ is measurable because of $f$ and has measure $\leq k \int |f|$ by Chebyshev, hence finite as $|f|$ is also integrable, and it's done.

2. We may write $B=\bigcap^{\infty}_{k=1}B_k$, where $B_k=\{x\mid |f(x)|\geq k\}$. The sequence $(B_k)$ is nonincreasing and $B_1$ is of finite measure again by Chebyshev, so we have $m(B)=\lim m(B_k)$. We have $0\leq m(B_k)\leq 1/k . \int |f|$ again using Chebyshev, so that by squeeze limit and as $|f|$ is integrable, we have $B$ as a null set.

Thanks for any suggestion. I wonder if there is another method especially if the above are incorrect.