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In my textbook we work with the following definition for continuous random variables:

A random variable X is continuous if its distribution function $F_X$ may be written in the form

$F_X(x)=\mathbb P(X\leq x)=\int_{-\infty}^{x}f_X(u)\text{d}u\quad \text{for } x\in \mathbb R,$

for some non-negative function $f_X$.

I'm wondering if $F_X$ is continuous, and if so, how to prove that.

The way I started:

We can write $\int_{-\infty}^{x}f_X(u)\text{ d}u=\lim_{a\to-\infty}\int_{a}^{x}f_X(u)\text{ d}u$. By the fundamental theorem of calculus, we know that $F^{\text{~}}(x)=\int_{a}^{x}f_X(u)\text{ d}u$ is continuous for each $a\in \mathbb R$. But how can I extend this to the case of the limit?

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    $F_X(x)$ is differentiable and therefore continuous. – 2017-01-21
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    How do we know $F_X(x)$ is differentiable. If $f_X$ is continuous, then of course $F_X$ is differentiable, but as far as I'm concerned, it's nowhere stated that $f_X$ is continuous. – 2017-01-21
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    You have defined it to be the anti-derivative of the frequency distribution. Therefore its derivative is the frequency distribution. For example. suppose $f(x)$ is the uniform distribution function over [0,1]. $f(x)$ is discontinuous at $0$ and at $1.$ Nonetheless, $F_x$ is continuous. – 2017-01-21
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    @DougM That is not correct. The function $f$ might be non-continuous, which would make $F_X$ non-differentiable (not everywhere, anyway). For example take for $f$ the indicator function for the unit interval... – 2017-01-21
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    @DanielRobert-Nicoud but $F_X$ still continuous in that example. I suppose we need additional information that $f(x)$ is bounded. So that the $\lim_\limits{x\to a-} \int_{-\infty}^a f(x) dx + \lim_\limits{x\to a+} \int_{a}^{x} f(x) dx$ converge, and thus we are able to "jump over" any discontinuities. – 2017-01-21
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    The relevant property is absolute continuity. Since $f_X$ is integrable, the measure $\mu(A) = \int_A f_x$ is absolutely continuous. – 2017-01-21
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    In fact, it is differentiable almost everywhere, and at such points, $F_X'(x) = f_X(x)$. – 2017-01-21

2 Answers 2

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$F$ is non-decreasing, so for each $x$ the one-sided limits $F(x_+)$ and $F(x_-)$ exist.

$F(x_-)=\int_{(-\infty,x)}f(x)dx$

By monotone convergence theorem:

$F(x_-)=\lim_{n\rightarrow\infty}F(x-1/n)=\lim_{n\rightarrow\infty}\int_{(-\infty,x-1/n]}f(x)=\int_{(-\infty,x)}f(x)dx$

$F(x)=\int_{(-\infty,x]}f(x)dx$

$F$ is right-continuous so $F(x)=F(x_+)$, regardless of its overall continuity:

$F(x)=P(X\le x)=P(\bigcap X\le x+1/n)=\lim_{n\rightarrow\infty}P(X\le x+1/n)=\lim_{n\rightarrow\infty}F(x+1/n)=F(x_+)$

So obviously $F$ is continuous at $x$, since the value of $f$ in one point has no effect on the integral.

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    "since the value of f in one point has no effect on the integral." why? – 2017-01-21
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    because a single point has Lebesgue measure zero – 2017-01-21
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    right, so we are using that the integral over a space of measure $0$ is $0$? Ok, that's easy to prove. But how do you use the fact that $F(x_)=F(x)$ to prove $F$ is continuous? – 2017-01-21
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    $F$ is right-continuous, and if we proved it is left-continuous that's it. – 2017-01-21
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    well, you didn't really prove that $F$ is left continuous, you proved that the integral over the set $(-\infty, x]$ is the same as the integral over the set $(-\infty,x)$. – 2017-01-21
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    OK, i added more info. – 2017-01-21
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It is continuous. To prove it we use two results:

  • Convergence of measure: If $\nu$ is a measure $\mu$ on a space $(X,\sigma)$ and $A_1\subseteq A_2\dots $ and $B_1\supseteq B_2,\dots $ are sets such that their union/intersection is equal to $C$ then we have $\lim\limits_{n\to \infty} A_n=\nu(C)$ and we have $\lim\limits_{n\to \infty} \nu(B_n)=\nu(C)$.
  • For any integrable function $f:X\rightarrow \mathbb R$ we have that the function $F(A)=\int_A fd\mu$ is a measure on $(X,\sigma)$.

Suppose you want to prove $F$ is continuous at $x$. It suffices to show that if we pick a monotone sequence $x_1,x_2,\dots$ converging to $x$ the $\lim\limits_{n\to \infty} F(x_n)=\lim\limits_{n\to \infty}\int\limits_{(-\infty,x_n]}fd\mu=\int\limits_{(-\infty,x)} fd\mu=\int \limits_{(-\infty,x]} f d\mu=F_(x)$. Which is exactly convergence of measures. The second to last equality is because the set $\{x\}$ has measure zero, and integrals over sets of measure zero are zero.

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    But how do I know for sure that $F(A)=\int_A fd\mu$ is a measure? I can follow the rest of your proof, so if I know how to show it's a measure, I'm there. – 2017-01-21
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    Oh wait, that's an assumption in my case! $F_X(x)=\mathbb P(X\leq x)$, so we already have that $F_X$ is a probability measure. I think I got the proof now, by approaching some $x\in \mathbb R$ from both sides, and using the continuity of $\mathbb P$ (which you call the convergence of the measure). – 2017-01-21
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    uhmm, I guess, althought we should be able to prove it without that, we did it in my measure theory class but I didn't pay that much attention :( – 2017-01-21
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    No problem, I'll ask my teacher anyway! Thanks for your help. – 2017-01-21
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    Happy to help. Im sure someone more profficient in the topic will drop some knowledge later. – 2017-01-21