The Problem: Let $X$ be a random variable with density function $f$, and assume that $f$ is continuous at $x=a$. Prove that $$\lim_{\varepsilon\searrow0}\frac{1}{\varepsilon}P(X\in(a,a+\varepsilon))=f(a).$$ My Attempt: Let $\eta>0$ be given. Since $f$ is continuous at $a$, it also right-continuous at $a$. Therefore, we can choose a $\delta>0$ such that if $0<\varepsilon<\delta$ then $\vert f(a+\varepsilon)-f(a)\vert<\eta.$ This also implies that if $0<x-a<\delta$ then $\vert f(x)-f(a)\vert<\eta.$ It follows that if $0<\varepsilon<\eta$, then \begin{align} \left\vert\frac{1}{\varepsilon}P(X\in(a,a+\varepsilon))-f(a)\right\vert&=\left\vert\frac{1}{\varepsilon}\int_a^{a+\varepsilon}[f(x)-f(a)\,dx\right\vert\\ &\leq\frac{1}{\varepsilon}\int_a^{a+\varepsilon}\vert f(x)-f(a)\vert\,dx\\ &<\frac{1}{\varepsilon}\int_a^{a+\varepsilon}\eta\,dx\\ &=\eta. \end{align} Therefore, $$\lim_{\varepsilon\searrow0}\frac{1}{\varepsilon}P(X\in(a,a+\varepsilon))=f(a).$$
Do you agree with my approach and execution above?
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