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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 $\mu(B)<\mu(A)$.

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    @joe: See the edit to my answer.2011-03-04

1 Answers 1

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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 $\mu(B) \gt 0$, then since $A$ is atomic, we have $\mu(A) = \mu(B)$ 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)$

Thus we must have that $\mu(B_1) \lt \mu(A)$ and so, $\mu(B_1) = 0$.

Similarly $\mu(B_2) = 0$ where $B_2 = A \cap \{x: f(x) \lt k\}$

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    Ahh, and similarly for $B_2$. Thank you so much!2011-03-03