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3.1 $\operatorname{\mathsf{P}}\left({X=x}\right)=F_{X}(x)-\lim_{i\rightarrow\infty}% F_{X}(x-1/i)$

The basic argument is given below and is relatively straightforward. However, it requires that

\displaystyle\operatorname{\mathsf{P}}\left({\cap_{i=1}^{\infty}\{\omega:x-1/i% <X(\omega)\leq x\}}\right)=\lim_{i\rightarrow\infty}\operatorname{\mathsf{P}}% \left({\{\omega:x-1/i<X(\omega)\leq x\}}\right),

which itself requires the axiom of ‘countable subadditivity’ and is proved in the subsection that follows.

Firstly, $\{x\}=\cap_{i=1}^{\infty}\{s:x-1/i<s\leq x\}$ , so

	$\displaystyle\operatorname{\mathsf{P}}\left({X=x}\right)$	$\displaystyle=\operatorname{\mathsf{P}}\left({\omega:X(\omega)=x}\right)$
		$\displaystyle=\operatorname{\mathsf{P}}\left({\cap_{i=1}^{\infty}\{\omega:x-1/% i<X(\omega)\leq x\}}\right)$
		$\displaystyle=\lim_{i\rightarrow\infty}\operatorname{\mathsf{P}}\left({\{% \omega:x-1/i<X(\omega)\leq x\}}\right)$

(see Appendix 3.1.1). So

	$\displaystyle\operatorname{\mathsf{P}}\left({X=x}\right)$	$\displaystyle=\lim_{i\rightarrow\infty}\operatorname{\mathsf{P}}\left({\{% \omega:X(\omega)\leq x\}}\right)-\operatorname{\mathsf{P}}\left({\{\omega:X(% \omega)\leq x-1/i\}}\right)$
		$\displaystyle=\lim_{i\rightarrow\infty}\operatorname{\mathsf{P}}\left({X\leq x% }\right)-\operatorname{\mathsf{P}}\left({X\leq x-1/i}\right)$
		$\displaystyle=\lim_{i\rightarrow\infty}F_{X}(x)-F_{X}(x-1/i)$
		$\displaystyle=F_{X}(x)-\lim_{i\rightarrow\infty}F_{X}(x-1/i).$

3.1 𝖯⁡(X=x)=FX⁢(x)-limi→∞⁡FX⁢(x-1/i)