A measurable function on an atom is almost everywhere constant Let $f \in m(\Omega,\mathcal{F})$, i.e. $f \mapsto [-\infty,\infty]$ and let $A \in \mathcal{F}$ be an atom. Prove that $f$ is almost everywhere constant on A: there exists $k \in [-\infty,\infty]$ such that 
$\mu (\{\omega \in A : f(\omega) \neq k \} )=0$.
I was thinking let $k=\frac{1}{\mu(A)}\int \limits_{A} f\,d\mu$.
Then let $B=\{\omega \in A :f(\omega) \neq k \}$.  Since A is an atom, and B is a subset of A then $\mu(B)=0$, in which case we're done, or $\mu(B)=\mu(A)$.  So I need to show that $\mu(B)<\mu(A)$.
 A: Looks like you almost have it: Try splitting $B$ into two sets.
More details:
Define $k$ as you did: $k\mu(A) = \int_{A} f \text{d}\mu$.
Let $B_1 = A \cap \{x: f(x) \gt k\}$
If $\mu(B_1) \gt 0$, then since $A$ is atomic, we have $\mu(A) = \mu(B_1)$ and thus
$ k\mu(A) = \int_{A} f \text{d}\mu = \int_{B_1} f \text{d}\mu \gt \int_{B_1} k\ \text{d}\mu = \int_{A} k\ \text{d}\mu = k \mu(A)$,
a contradiction. Thus we must have that $\mu(B_1) \lt \mu(A)$ and so, $\mu(B_1) = 0$.
Similarly $\mu(B_2) = 0$ where $B_2 = A \cap \{x: f(x) \lt k\}$
A: In your post and the answer provided we are making two assumptions : $f$ is integrable and $\mu(A)<\infty$
This post provides a counterexample to your claim : https://mathoverflow.net/questions/149936/are-measurable-functions-almost-surely-constant-on-atoms
So let us consider only integrable functions. Then only the case $\mu(A)=\infty$ remains. WLOG assume that $\int\limits_A f\geq 0$. If $f=0$ a.e. in $A$ then we are done. If not, then $f\neq 0$ a.e. in $A$ (since atom). Then $\{x\in A: f(x)<0\}$ has measure $0$ so that $f>0$ a.e. in $A$. But $$\{x\in A:f(x)>0\}=\bigcup\limits_n\{x\in A:f(x)>\frac{1}{n}\}$$ So $f>\frac{1}{n}$ a.e. in $A$ for some $n$. But then $$\int\limits_A f>\frac{1}{n}\mu(A)=\infty$$ yields a contradiction.
