Proof that CDF is continuous for continuous random variables.

Question

I have to show that,

For any continuous random variable $X$, the cumulative distribution function(CDF) $F : \mathbb{R} \to [0,1]$ is continuous.

My attempt

Assume $F$ has a discontinuous point $x \in \mathbb{R}$. Since CDF should be non-decreasing, i.e. monotone increasing, so it follows that $f(x-) < f(x+)$. Since CDF is right-continuous, $f(x)=f(x+)$, thus $X$ has point mass of size $f(x+) - f(x-)$ at the point $x$. This contradicts the fact that $P(X=x) = 0$ for any $x \in \mathbb{R}$, because $X$ : continuous random variable.

Is my proof OK?

Answer 1

If you are allowed to assume $P(X=x)=0$, you're proof is fine, but how would you proof $P(X=x)=0$? Actually you are trying to proof that the CDF has no jumps, and you use the fact that the CDF has no jumps.

I think it would be best to note that a random variable is continuous when it has a probability density function $f$, and we can write $P(X\leq x)=\int_{-\infty}^xf(u)\mathrm{d}u$. Continuity of the continuous random variable now follows directly from the fact the Riemann integrals are continuous.

If you want to proof that $P(X=x)=0$: $$P(X=x)=\lim_{\epsilon\to0^+}P(x-\epsilon<X\leq x)=\lim_{\epsilon\to0^+}[F_X(x)-F_X(x-\epsilon)]=\lim_{\epsilon\to0^+}\int_{x-\epsilon}^xf_X(u)\mathrm{d}u=0,$$ where we use the continuity of the probability measure in the first equality.