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I'm preparing to the second mini-test in measure theory. Here is one of the problems I cannot deal with. I would appreciate any help, thank you.

Let $\mu$ be a Radon measure on $\mathbb{R}$, suppose that $A$ is a $\mu$–measurable subset of $[a,b]$ and let $h$ be a positive number. Prove that $$\frac{1}{2h}\int_a^b\mu(A\cap(x-h,x+h))\,\text{d}x\le \mu(A).$$

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If you apply Fubini (or Tonelli) in the second equality below (everything is positive) you get $$ \frac1{2h}\,\int_{[a,b]}\mu(A\cap(x-h,x+h))\,dx=\frac1{2h}\,\int_{[a,b]}\int_{(x-h,x+h)}\,1_A(t)\,d\mu(t)\,dx\\ \leq\frac1{2h}\int_{(a-h,b+h)}\int_{[t-h,t+h]}\,1_A(t)\,dx\,d\mu(t)=\int_{(a-h,b+h)}\,1_A(t)\,d\mu(t)=\mu(A\cap(a-h,b+h))=\mu(A). $$

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    So do we have in fact the equality?2012-11-19
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    No, we don't. Now I see where my mistake was. When I'm doing Tonelli, the region on the right-hand-side is smaller than the one I'm using. So I'm changing that equal sign to a $\leq$. If that's not clear enough, let me know and I'll try to clarify.2012-11-19
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    Do you mean the reason for inequality sign is that you enlarge the region from $[a,b]$ to $(a-h,b+h)$?2012-11-19
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    No. The region of integration is a parallelogram. When you swap the variables for Tonelli, its expression complicates a little. If I have to explain this to my Calculus II students, I would say that the region (in Stewart's Calculus language) is Type I but not Type II. So, for the second integrals I'm switching to a slightly larger parallelogram, hence the inequality. If you draw the region, you'll see it quickly.2012-11-19
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    Ah, ok, now I see. We are adding two small triangles to the initial parallelogram and integrating over slightly larger one. All is clear now, thank you for your help!2012-11-19
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    Exactly. You are welcome!2012-11-19