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Let $f$ be a distribution function (i.e. non-decreasing an right-continuous) on the real line.

In this question, for example, it is proved, that the set $D$ of points of discontinuity of $f$ is then at most countable.

I have the following problem/question with the proof from the link (or any other proof that I have seen of this statement): If $f$ is right-continuous we know that for every $x_0$ the limit $\lim_{x \rightarrow x_0, \ x\geq x_0} f(x)$ exists and is equal to $f(x_0)$. But, as far as I can see, we don't know that the limit from the left, $\lim_{x \rightarrow x_0, \ x< x_0} f(x)$, or as it is denoted in the link I gave, $f(x-)$, has to exist! But all the proofs assume this existence. So how can I prove the existence ?

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Hint: Show that indeed $f(x-)$ exists by identifying it: $$ f(x-)=\sup\limits_{t\lt x}f(t). $$

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If $f : [a,b] \rightarrow \mathbb{R} $ is a monotonic function then $f(x-)$ and $f(x+)$ exist in every point and $$ \sup_{a<t<x} f(t) = f(x-) \leq f(x) \leq f(x+) = \inf_{x<t<b} f(t) $$

A proof of this fact can be obtained combining the least upper bound property, its definition and the monotonicity of the function.