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The question is

$F$ is a distribution function. Show that $$P\{X = x\} = F(x) - \lim_{y \uparrow x}F(y)$$

This is trivial. But how can we prove it in a formal way?

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$F(y) = \Pr (X \le y)$ so $$\lim_{y \uparrow x}F(y) =\lim_{y \uparrow x} \Pr (X \le y) = \Pr (X \lt x) $$ and thus $$F(x) - \lim_{y \uparrow x}F(y)= \Pr (X \le x)- \Pr (X \lt x)= \Pr (X = x).$$

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The key point is that $\text{Pr}(X < x) = \lim_{y \uparrow x} \text{Pr}(X \le y)$ because of countable additivity: $$\{X < x\} = \bigcup_{n=1}^\infty \{X \le x - 1/n\} = \{ X \le x - 1\} \cup \bigcup_{n=2}^\infty \{x-1/(n-1) < X \le x - 1/n\}$$ so $$\text{Pr}(X < x) = \text{Pr}(X \le x - 1) + \sum_{n=2}^\infty \text{Pr}(x-1/(n-1) < X \le x - 1/n) = F(x-1) + \sum_{n=2}^\infty \left(F(x-1/n) - F(x-1/(n-1))\right) = \lim_{n \to \infty} F(x - 1/n)$$ –  Robert Israel May 6 '12 at 21:35
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@Henry : When you write 5\lim_{y\to x} g(y) in a "displayed" (as opposed to "inline") setting, then it looks like this: $\displaystyle5\lim_{y\to x} g(y)$ with "$y\to x$" directly below "$\lim$", proper spacing before and after "$\lim$", and "$\lim$" not italicized. That is standard. But when you write 5 lim_{y\to x} g(y), with no backslash in \lim, then it looks like this: $\displaystyle 5 lim_{y\to x} g(y)$. (In an "inline" setting, the subscript $y\to x$ is not directly below "$\lim$" unless you use \displaystyle.) –  Michael Hardy May 6 '12 at 22:02

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