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Let $0

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    what did you try? – 2017-01-01

1 Answers 1

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you can build it by parts, Just make sure that the measure of the set in the intervals $(p+\frac{1}{n+1},p+\frac{1}{n})$ and $(p-\frac{1}{n+1},p-\frac{1}{n})$ is equal to $a(\frac{1}{n}-\frac{1}{n+1})=\frac{a}{n(n+1})$.

Since $\lim\frac{1}{n(n+1)}/{\frac{1}{n+1}}=0$ this set will satisfy $\lim_{\delta \to 0}\frac{1}{2\delta}\int_{p+\delta}^{p-\delta}\chi_E(x)dx=a$

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    Thank for your answer. But I dont know why we should chose two intervals $(p+\frac{1}{n+1},p+\frac{1}{n})$ and $(p-\frac{1}{n+1},p-\frac{1}{n})$ is equal to $a(\frac{1}{n}-\frac{1}{n+1})=\frac{a}{n(n+1})$. If I let the measure of the set $(p-\frac{1}{n},p+\frac{1}{n})$ is equal to a(\frac{2}{n}) , it will also right ? – 2017-01-01
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    No, because in this construction the measure of the part of the set inside the interval is $a$ multiplied by the length of the interval, which is what yo want. – 2017-01-01
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    But if I choose the interval $(p-\frac{1}{n},p+\frac{1}{n})$ and it measure is $a(\frac{2}{n})$ . I think it also satisfy $\lim_{\delta \to 0}\frac{1}{2\delta}\int_{p+\delta}^{p-\delta}\chi_E(x)dx=a$. Is there have anything wrong but I dont find ? – 2017-01-01