Let $X$ have cdf $F(x)=(x/2+1/2)I_{[0,1)}(x)+I_{[1,\infty)}(x)$. Clearly, $P(X=0)=1/2$, so $X$ is a mixed random variable (is not discrete nor continuous). I'm asked to show that there is no function $f(x)$ such that $P(B)=\sum_{x_i\in B}f(x_i)$ for all Borel set $B$.
Is it sufficient to say that $B=(0,1)$ is not countable? I mean, you can not obtain $P(B)$ via $\sum_{x_i\in B}f(x_i)$ for some function $f$.
Best Answer
I see why you are confused and I think your idea is correct, but maybe formalise it. First, note that there is a general definition of an uncountable series given by $$ \sum_{x\in B} f(x) := \sup_{A\subset B, |A|<\infty} \sum_{x\in A} f(x). $$ Now assume that such an $f$ exists. Then $P((0,1))=1/2$, which means that there exists a sequence of finite sets $(A_n)_n$ in $(0,1)$ such that $$ \lim_{n\rightarrow\infty} \sum_{x\in A_n} f(x)=\frac{1}{2}. $$ In particular there exists a $\widehat{x}\in (0,1)$ with $f(\widehat{x})>0$ which is a contradiction to $$ P(\{\widehat{x}\})= F(\widehat{x})- \lim_{x\nearrow \widehat{x}} F(x)=0. $$