Is $\lim\limits_{n \to \infty} f(x_{0} + \frac{1}{n}) = \lim\limits_{h \to 0^{+}} f(x_{0} + h) \triangleq f(x_{0}^{+})$?

Question

We know in probability, the cumulative distribution function is right continuous. The proof of this statement used a relation that $$\lim\limits_{n \to \infty} F_{X}\left(x_{0} + \frac{1}{n}\right) = \lim\limits_{h \to 0^{+}} F_{X}(x_{0} + h) \triangleq F_{X}(x_{0}^{+})$$ However, I do not know why this is true? Or is it true in general or only for CDF's?

Answer 1

It is not true in general. For CDFs it is true because \begin{align} \lim_{n\to \infty} F_X\left(x+\frac{1}{n}\right)&=\lim_{n\to \infty}\mathrm{Pr}\left(X \le x+\frac{1}{n}\right)\\&=\mathrm{Pr}\left(\bigcap_n (X\le x+\frac{1}{n})\right)\\ & =\mathrm{Pr}\left(X\le x\right)=F_X(x). \end{align}