Let $F$ be a cumulative density function on $\mathbb{R}$. From an argument in a textbook, it is shown that $F$ must be right-continuous:

Let $x$ be a real number and let $y_1$, $y_2$, $\ldots$ be a sequence of real numbers such that $y_1 > y_2 > \ldots$ and $\lim_i y_i = x$. Let $A_i = (-\infty, y_i]$ and let $A = (- \infty, x]$. Note that $A = \cap_{i=1}^\infty A_i$ and also note that $A_1 \supset A_2 \supset \ldots$. Because the events are monotone, $\lim_i P(A_i) = P(\cap_i A_i)$. Thus,

$$ F(x) = P(A) = P( \cap_i A_i) = \lim_i P(A_i) = \lim_i F(y_i) = F(x^+) $$

But why doesn't this argument work in reverse to show that $F$ is left-continuous? That is, if we supposed that the $y_i$ were approaching $x$ from the left, why can't we analogously say:

$$ F(x) = P(A) = P( \cup_i A_i) = \lim_i P(A_i) = \lim_i F(y_i) = F(x^-)? $$

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    $\begingroup$ Because $\cup_{i=1}^\infty A_i$ would be $(-\infty, x)$, not $(-\infty,x]$. If you don't see it, which term of $\cup_{i=1}^\infty A_i$ contains $x$? $\endgroup$ – Erick Wong May 31 '16 at 16:08
  • $\begingroup$ Ok, so we could just adjust the stipulation to be that $A_i = (-\infty, y_i]$, could we not? $\endgroup$ – user1770201 May 31 '16 at 16:27
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    $\begingroup$ $A_i=(-\infty,y_i]$, and $y_i$ increases to $x$, which means each $y_i$ is less than $x$, so none of the $A_i$ contain $x$. $\endgroup$ – Ian May 31 '16 at 16:28
  • $\begingroup$ Yes -- you're right. So the mistake in my proof is then $P(A) = P(\cup_i A_i)$. That fails to hold. The reasoning to the right is still valid though, correct? (Even though the equation as a whole doesn't hold)? $\endgroup$ – user1770201 May 31 '16 at 16:29
  • $\begingroup$ @user1770201 The reasoning leads to the accurate conclusion that the left-hand limit $F(x^-)$ is equal to $F(x) - P(X=x)$. $\endgroup$ – Erick Wong May 31 '16 at 16:48

The right continuity of CDFs is a matter of convention. One could just as well take the default to be the left-continuous function $x\mapsto \Bbb P[X<x]$, as one finds in the Russian literature of yesteryear (and perhaps even today); cf. B.V. Gnedenko's Theory of Probability.

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    $\begingroup$ It is more that the fact that $X=x$ is included in the event whose probability is $F(x)$ is a convention. The right-continuity (and possible lack of left-continuity) follows from that. But yes, there is no real loss in using the convention under which CDFs are left-continuous. $\endgroup$ – Ian May 31 '16 at 16:27

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