Prove that the CDF of a random variable is always right-continuous

Question

Let $X$ be a random variable with cumulative distribution function $F_X$. It is a known fact that this function $F_X$ is right-continuous. But I'm having some trouble to prove this result. Below I'm sketching out what I have done so far to prove this and where exactly I'm having the difficulty.

My attempt:

To show that $F_X$ is right-continuous at a real number $y$, we need to prove that $$\lim_{x\to y^+}F_X(x)=F_X(y).$$

I will use the sequential approach to prove this. Let $(x_n)$ be a sequence such that $x_n \to y$ and $x_n \geq y$ for every $n$.

Define $A_n = (-\infty,x_n]$ for each $n$. Now from the sequence $(x_n)$ we can extract a monotonically decreasing subsequence, say $(x_{n_k})$. As the new subsequence is monotonically decreasing, it follows from our definition of $A_n$'s that for all $k$ in $\mathbb{N}$, $A_{n_{k+1}} \subseteq A_{n_k}$. So $X^{-1}(A_{n_{k+1}}) \subseteq X^{-1}(A_{n_k})$ and hence $$F_X(x_{n_{k+1}}) = P(X\leq x_{n_{k+1}}) = P(X^{-1}(A_{n_{k+1}})) \leq P(X^{-1}(A_{n_{k}})) = P(X\leq x_{n_{k}}) = F_X(x_{n_{k}}).$$

Now since $(X^{-1}(A_{n_k}))_k$ is a nested sequence of events, we know $$\lim_{k\to\infty}P(X^{-1}(A_{n_k})) = P(\cap_{k=1}^{\infty}(X^{-1}(A_{n_k}))) = P(X^{-1}(\cap_{k=1}^{\infty}A_{n_k})) = P(X^{-1}(-\infty,y]).$$

Therefore, $$\lim_{k\to\infty}F_X(x_{n_k}) = P(X\leq y) = F_X(y).$$

This shows that the sequence $(F_X(x_{n_k}))_k$ converges to $F_X(y)$. But it does not prove $F_X(x_n) \to F_X(y)$ as $n\to \infty$, which is required to show that $F_X$ is right-continuous. I'm completely stuck at this point and can't get any idea how to go from the particular subsequence $(F_X(x_{n_k}))_k$ to the general sequence $(F_X(x_n))_n$ or how to prove this result using the sequential approach and without having to extract a monotone decreasing subsequence from a given sequence.

I'll appreciate any help or suggestion. Thanks and regards.

Answer 1

Your struggle comes from a lack of understanding of continuity. In order to show $f:\mathbb{R}\to\mathbb{R}$ is (left/right)-continuous, it suffices to show that if $x_n \to x$ monitonically (from the left/right) that $f(x_n) \to f(x)$.

This has nothing to do with probability, it follows from basic real analysis. Try to work out the proof. (It should only be 3-4 lines at most). Start from the definition of continuity.

Answer 2

To be able to properly talk about the notion of continuity for CDFs or PDFs, it is better to use measure theoretic foundation of probability.

But anyway, your argument can be completed as follows. Choose $A_n=\left(-\infty,x+\frac{1}{n}\right]$ as your intervals. As you said, you can prove $F_X\left(x+\frac{1}{n}\right) \to F_X(x)$. Now , for each $\epsilon$, you can find $n$ such that: $$0 \leq F_X\left(x+\frac{1}{n}\right)-F_X(x)<\epsilon.$$ Because the CDF is monotonic, choose $\delta<\frac{1}{n}$. Then for all $y$, with $x<y<x+\delta$ we have: $$0 \leq F_X(y)-F_X(x)\leq F_X(x+\delta)-F_X(x)\leq F_X\left(x+\frac{1}{n}\right)-F_X(x)<\epsilon,$$ which finishes the proof.