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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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btw: +1 for showing the working. –  Aryabhata Mar 3 '11 at 18:20
    
can you spell out the rest of it for me? –  user7760 Mar 3 '11 at 22:05
    
@joe: please do not use answers to make comments. –  Qiaochu Yuan Mar 3 '11 at 22:21
    
@joe: See the edit to my answer. –  Aryabhata Mar 4 '11 at 1:58

1 Answer 1

up vote 4 down vote accepted

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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So let $B = B_1 \cup B_2$. Then $\mu(B_1)=0$ or $\mu(A)$ and $\mu(B_2)=0$ or $\mu(A)$. Not sure what to do, maybe you meant split $B$ into specific sets? –  James Mar 3 '11 at 18:41
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@James: $B_1 = \{\omega : f(w) \gt k\}$ –  Aryabhata Mar 3 '11 at 18:42
    
OK and $B_2=\{\omega:f(\omega)<k\}$. Now if $\mu(B_1)=\mu(A)$ then $\mu(B_2)=0$ and our $k$ from the integral changes value? –  James Mar 3 '11 at 19:00
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@James: if $\mu(B_1) = \mu (A)$, then $\int_A f = \int_{B_1} f \gt \int_{B_1} k $... –  Aryabhata Mar 3 '11 at 19:07
    
Ahh, and similarly for $B_2$. Thank you so much! –  James Mar 3 '11 at 19:10

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