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How can I prove that if the density function $f_X$ of a (continuous) random variable $X$ is continuous, then cumulative distribution function $F(x)$ is also continuous?

Given the definition of the CDF, we have,

$F(x) = \int_{-\infty}^xf_X(u)du$

So one approach could be to prove the above function is continuous. There are proofs of continuity (in analysis) when the integral limits are bounded, but I'm not sure if there is something special about the lower bound of $-\infty$.

Also, I would like to find a proof that does not use results from measure theory.

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Not only is $\ F\ $ continuous, it's actually differentiable, which implies its continuity: \begin{align} \left|\,\frac{F(x+\delta x)-F(x)}{\delta x}-f_X(x)\,\right|&=\left|\frac{\int_x^{x+\delta x}\big(f_X(u)-f_X(x)\big)\,du}{\delta x}\,\right|\\ &\le\frac{\int_x^{x+\delta x}\big|f_X(u)-f_X(x)\big|\,du}{|\,\delta x\ |}\\ &\le\sup_{x-|\delta x|\le u\le x+|\delta x|}\big|f_X(u)-f_X(x)\big|\ . \end{align} Because $\ f_X\ $ is continuous, then $$ \lim_{\delta x\rightarrow0}\sup_{x-|\delta x|\le u\le x+|\delta x|}\big|f_X(u)-f_X(x)\big|=0\ . $$ Therefore, by the sandwich theorem, $$ \lim_{\delta x\rightarrow0}\frac{F(x+\delta x)-F(x)}{\delta x}=f_X(x)\ , $$ so $\ F\ $ is differentiable with derivative $\ f_X(x)\ $ at $\ x\ $.

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