Showing a distribution function is right but not left-continuous

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.

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In your 'proof' of left-continuity $\cap (y_n,x]$ is not empty. – user20266 Dec 11 '11 at 11:40
Sir, I am new to measure theory but I think the statement which says ..."F is defined as $\u$" – roni Aug 18 '14 at 12:29

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.
Let $y_{n}\uparrow x$ so $(y_{1},x]\supseteq(y_{2},x]\supseteq...$, and assume $a\in\cap_{n}(y_{n},x]$. Then $y_{n}<a\leq x$ for all $n$. Since $y_{n}\rightarrow x$, then for any $\epsilon>0$ there exists an $N$ such that $-\epsilon<y_{n}-x\leq 0$, for all $n>N$. Since $y_{n}<a$ for all $n$ implies $y_{n}-x<a-x$ for all $n$, we must have $-\epsilon<a-x\leq 0$. Thus we must have $a=x$. Now if we assume $a'\in\cap_{n}(y_{n},x]$ for $a'\not=a$ then we get the same result, i.e., $a'=x$. So $a'=a=x$ is the only real number common to all the intervals $(y_{n},x]$, and so $\cap_{n}(y_{n},x]=\{x\}$. – dandar Dec 13 '11 at 19:17
Dear user20266, why do you take intersection of the $(y_n, x]$. If we want to look at left continuity, we want to look at the intervals $(0, y_n)$ that are approaching $(0, x] = (0, \cup y_n]$. We do have union not intersection. no? – Ievgenii Jan 25 at 13:38