I am having difficulty proving right-continuity of a distribution function. My problem is that my method for showing right-continuity works also to show $F$ is left-continuous. My proof of both must therefore be wrong but I do not know where. A distribution function of a random variable $X$ can be defined as
$F(x)=\mu(-\infty,x]=P[X\leq x],$
where $P$ is the probability measure of the underlying probability space $(\Omega,\mathscr{F},P)$, and $\mu$ is a probability measure on the 1-dimensional Borel sets $\mathscr{R}$. Using the monotonicity of $\mu$ we see that $F$ is non-decreasing. Also for $a\leq x$ we have $\mu(a,x]=\mu(-\infty,x]-\mu(-\infty,a]$. I think the fact that $F$ is defined as $\mu(-\infty,x]$ and not $\mu(-\infty,x)$ means that $F$ can have a jump discontinuity say at $a$ as $x$ approaches from the left but not from the right – i.e $F$ is right but not left-continuous. My proofs rely on the following result: if $A_{n}$ and $A$ lie in $\mathscr{F}$ and $A_{n}\downarrow A$, and if $\mu(A_{1})<\infty$, then $\mu(A_{n})\downarrow \mu(A)$.
Here is my proof for right-continuity;
Let $y_{n}\downarrow x$. Then $(x,y_{n}]\downarrow \emptyset:(x,y_{1}]\supseteq(x,y_{2}]\supseteq…$ and $\cap_{n}(x,y_{n}]=(x,x]=\emptyset$. Since $(x,y_{n}]$ and $\emptyset$ are in $\mathscr{F}$, and $\mu(x,y_{1}]\leq 1<\infty$, then we have $\mu(x,y_{n}]\downarrow\mu(\emptyset)=0$. Equivalently $F(y_{n})-F(x)\downarrow 0:F(y_{1})-F(x)\geq F(y_{2})-F(x)\geq…$ and $F(y_{n})-F(x)\rightarrow 0$. Thus we have $F(y_{1})\geq F(y_{2})\geq …$, and $F(y_{n})\rightarrow F(x)$. So $F$ is right-continuous: $lim_{y\downarrow x}F(y)=F(lim_{y\downarrow x})$.
My "proof" for left-continuity is as follows;
Let $y_{n}\uparrow x$. Then $(y_{n},x]\downarrow\emptyset:(y_{1},x]\supseteq(y_{2},x]\supseteq…$ and $\cap_{n}(y_{n},x]=(x,x]=\emptyset$. Since $(y_{n},x]$ and $\emptyset$ are in $\mathscr{F}$, and since $(y_{1},x]\leq 1<\infty$ then $\mu(y_{n},x]\downarrow\mu(\emptyset)=0$, or in terms of $F$ we have $F(x)-F(y_{n})\downarrow 0$. Thus $F(x)-F(y_{1})\geq F(x)-F(y_{2})\geq…$ and $F(x)-F(y_{n})\rightarrow 0$ or equivalently $F(y_{n})\uparrow F(x):F(y_{1})\leq F(y_{2})\leq…$ and $F(y_{n})\rightarrow F(x)$. So $F$ is left-continuous: $lim_{y\uparrow x}F(y)=F(lim_{y\uparrow x})$.
Where am I going wrong? Any help would be greatly appreciated.
Best Answer
I could as well write an answer instead of a comment :-)
if $y_n \rightarrow x$ is increasing and $y_n \neq x, \forall n$ then $\cap_n (y_n,x] = \{x\}$, which may have positive measure.