Exercise 0.26 (Properties of Stieltjes measure function)

Question

Let $(\mathbb{R},\mathcal{B},P)$ be the real probability space, and let $F$ be the Stieltjes measure function associated with $P$. Show that\begin{enumerate}\item $F$ is non-decreasing.\item $\lim_{t	o-\infty}F(t) = 0$ and $\lim_{t	o+\infty}F(t) = 1$.\item $F$ is right continuous, i.e., for all $t\in\mathbb{R}$ we have $F(t) = \lim_{s	o t^+} F(s)$.
ewline\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}
	\item If $y \geq x$, then $(-\infty,x] \subseteq (-\infty,y]$, so $P\left((-\infty,x]ight) \leq P\left((-\infty,x]ight)$ by the monotonicity property of the probability measure $P$ (\href{./exercise_23}{Exercise 23}). Thus, we have $F(x) \leq F(y)$.

\item By \href{../analysis_I/exercise_9-3-1}{Exercise 9.3.1} of \href{../analysis_I}{Analysis I} we can replace the continuous $\lim_{t	o -\infty} F(t)$ by the convergence along any arbitrary sequence $(t_n)_{n\in\mathbb{N}}$ in $\mathbb{R}$ with $t_n\geq t$. (Without loss of generality, we can make the sequence monotone decreasing also by setting $t_n' = t_i$ such that $t_i$ is the first element with property $t_i\geq t_{n-1}'$ and $i>n-1$.) But then $((-\infty,t_n])_{n\in\mathbb{N}}$ is a sequence of nested $\mathcal{B}$-measurable sets and so we can use the \href{./exercise_23}{downwards monotone convergence (Exercise 23)} to deduce that$$\lim_{n	o\infty} F(t_n) = \lim_{n	o\infty} P((-\infty,t_n]) = P\left( \bigcap_{n=1}^{\infty} E_night) = P(\emptyset) = 0$$Similarly, by the \href{./exercise_23}{upwards monotone convergence (Exercise 23)} we see that$$\lim_{n	o\infty} F(t_n) = \lim_{n	o\infty} P((-\infty,t_n]) = P\left( \bigcup_{n=1}^{\infty} E_night) = P(\mathbb{R}) = 1.$$

\item Again, we use a monotone decreasing sequence $(s_n)_{n\in\mathbb{N}}$ with $s_n \geq t$. We have two cases:
		\begin{itemize}
			\item $(s_n)_{n\in\mathbb{N}}$ eventually becomes $t$, in which case the theorem assertion follows obviously
			\item We have eventually $s_n 
eq t$ for all $n$.
		\end{itemize}
		We inspect the latter case. By additivity of the probability measure combined with the \href{./exercise_23}{downwards monotone convergence (Exercise 23)} we obtain:
		\begin{align*} \lim_{n	o\infty}F(s_n) &= \lim_{n	o\infty} P\left((-\infty,s_n]ight)\ &= \lim_{n	o\infty} P\left((-\infty,t]ight) + P\left((t,s_n]ight)\ &= P\left((-\infty,t]ight) + \lim_{n	o\infty} P\left((t,s_n]ight)\ &= F(t) + P\left(\bigcap_{n\in\mathbb{N}} (t,s_n]ight)\ &= F(t) + P(\emptyset) = F(t)\end{align*}
\end{enumerate}

\begin{remark}
Note that this trick will not work with the left limits since
\begin{align*} \lim_{n	o\infty}F(s_n) &= \lim_{n	o\infty} P\left((-\infty,s_n]ight)\
&= \lim_{n	o\infty} P\left((-\infty,t]ight) - P\left((s_n,t]ight)\
&= P\left((-\infty,t]ight) - \lim_{n	o\infty} P\left((s_n,t]ight)\
&= F(t) - P\left(\bigcap_{n\in\mathbb{N}} (s_n,t]ight)\ &= F(t) - P(\{t\}).
\end{align*}
\end{remark}

Exercise 0.26 (Properties of Stieltjes measure function)

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Exercise 0.26 (Properties of Stieltjes measure function)

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