Unique Measure from Right Continuous and Left Continuous CDF

Question

Consider $\mathbb R$ equipped with the usual Borel $\sigma$-algebra.

For each probability measure $P$ on $\mathcal B$ there exists a unique monotone, right-continuous function $F$ on $\mathbb R$ satisfying $\lim_{x\rightarrow\infty} F(x) = 1$ and $\lim_{x\rightarrow-\infty}F(x) = 0$. Conversely, for each monotone, right-continuous function $F$ on $\mathbb R$ satisfying $\lim_{x\rightarrow\infty} F(x) = 1$ and $\lim_{x\rightarrow-\infty}F(x) = 0$ there exists a unique probability measure $P$ on $\mathcal B$.

The proof for the first part defines $F(x) = P((-\infty,x])$ and shows that this function the required properties. For the second part, one defines $\mu((a,b]) := F(b) - F(a)$ and then shows that $\mu$ is a $\sigma$-finite pre-measure on the algebra of left half-open intervals. Caratheodorys extension theorem then yields a unique measure $P$ that coincides with $\mu$ on the algebra of half-open intervals.

The above procedure also works, mutatis mutandis, for left-continuous functions. Let $F^+$ and $F^-$ denote the respective right- and left-continuous functions corresponding to some probability measure $P$ on $\mathcal B$. Next, define a new function $G$ by $$G = \frac{F^+ + F^-}2.$$ Does $G$ correspond to a unique probability measure $Q$ on $\mathcal B$?

My approach: I tried to mimic the proof for the right/left-continuous case. So I defined $$\mu((a,b)) _= \frac{F^+(b) + F^+(b)}2 - \frac{F^+(a) + F^-(a)}2.$$ But the set of open intervals is not an algebra, which suggests that this idea does not work. Since we know that both $F^+$ and $F^-$ each induce a probability measure $P^+$ and $P^-$, we could simply define $P := \frac{P^+ + P^-}2$ to get a probability measure. However, is $P$ unique?

Answer 1

$G(x)=0$ for all but countable many $x$ and, by monotonicity, $G(x)=0$ for all $x$. So the answer is NO.

[If you meant to write $G = \frac{F^{+} +F^{-}}2$ then uniqueness follows from the fact that $G=F$ at all but countably many points and $P(a,b)$ is uniquely determined whenever $a$ and $b$ are continuity points of $F$].

Answer 2

$G=0$ corresponds to any distribution of a continuous random variable so no, $G$ does not correspond to a unique probability measure.

Answer 3

"Does $G$ correspond to a unique probability measure $Q$ on $\mathcal B$?" That depends on what you mean by "corresponds."

Given a probability measure $P$ in the Borel sets $\mathcal B$, you have the right continuous cdf $F^+(x):=P\left((-\infty,x]\right)$, $x\in \Bbb R$.

Conversely, given a right continuous non-decreasing function $\Phi^+$ with limits $0$ and $1$ at $-\infty$ and $+\infty$, there is a unique probability measure $\Pi_+$ on $\mathcal B$ such that $\Pi_+\left((-\infty,x]\right)=\Phi^+(x)$ for all real $x$.

There is also a left continuous function $F^-$ determined by $P$ via the recipe $F^-(x) := P\left((-\infty,x)\right)$, $x\in\Bbb R$. Of course, $F^+(x-) =F^-(x)$ and $F^-(x+)=F^+(x)$ for all $x$. (Here $F^+(x-)$ denotes the left limit of $F^+$ at $x$, etc.) Also, $F^+(x) -F^-(x) = P\left(\{x\}\right)$ for all $x$.

Conversely, given a left-continuous non-decreasing function $\Phi^-$ with limits $0$ and $1$ at $-\infty$ and $+\infty$, there is a unique probability measure $\Pi_-$ on $\mathcal B$ such that $\Pi_-\left((-\infty,x)\right)=\Phi^-(x)$ for all real $x$.

Finally, if $\Phi^+$ and $\Phi^-$ are right- (resp. left-) continuous non-decreasing functions with the appropriate limits at $\pm\infty$, such that $\Phi^+(x-) = \Phi^-(x)$ for all $x$, then the associated probability measures $\Pi_+$ and $\Pi_-$ are the same.

There is no special probability measure associated with your $G$. Rather, $$ G(x) = P\left((-\infty,x)\right)+ {1\over 2}P\left(\{x\}\right) = P\left((-\infty,x]\right)- {1\over 2}P\left(\{x\}\right),\qquad\forall x\in\Bbb R. $$