# Prove that $\frac{1}{2h}\int_a^b\mu(A\cap(x-h,x+h))\,\text{d}x\le \mu(A)$

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? – Kuba Helsztyński Nov 19 '12 at 0:26
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. – Martin Argerami Nov 19 '12 at 0:32
Do you mean the reason for inequality sign is that you enlarge the region from $[a,b]$ to $(a-h,b+h)$? – Kuba Helsztyński Nov 19 '12 at 0:37
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. – Martin Argerami Nov 19 '12 at 0:40
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! – Kuba Helsztyński Nov 19 '12 at 1:22