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Let $\mu$ be the Lebesgue measure over $\mathbb{R}$ and $A\subset\mathbb{R}$ such >that $\mu(A)<+\infty$. Demonstrate that the function $$f(x) := \mu(A\cap(-\infty,x])$$

is continuous.

I was trying to do the demonstration but I’m not sure how to modify the sets to use the hypothesis of $\mu(A)$.

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Hint: use the continuity of the measure to show that the sided limits of the function are equal for every $x$.

you have to show that for every $\epsilon > 0$ there is a $\delta > 0$ such that if $|y – x| < \delta$ then $|f(x) – f(y)| < \epsilon$.

Let $y < x$: $$|f(x) – f(y)| = \mu(A \cap (-\infty,x]) – \mu(A \cap (-\infty,y]) = \mu(A \cap (y,x])$$Note that $A \cap (y,x] \subset (y,x]$. This implies that $\mu(A \cap (y,x]) \le \mu((y,x]) \le \delta$, since the Lebesgue measure of any interval equals its legth (endpoints are irrelevant to this computation!) Then choice $\delta < \epsilon$ to conclude in this case. The case when $y \ge x$ is symmetric, hence the proof is complete.

I hope this helps!!!

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