Sum of random 0 and 1 Let $x_i$ be uniformly distributed random variables in the interval $[0,a]$, with $a>0$.
Let f(x) be equal to 1 if x=0, and 0 otherwise.
Let $$S(a)=\sum_{n=1}^\infty f(x_n)$$
What is S(1)?
What is $\lim_{a->0} S(a)$ from positive side?
 A: For any fixed $a > 0$,
$$
\renewcommand{\Pr}{\mathbb{P}}
\Pr(S(a) = 0) = \Pr\Big(\sum_{n=1}^\infty f(X_n) = 0\Big) = \Pr(\cap_n \{X_n > 0\}) = 1
$$
where the last line follows since $\Pr(X_n > 0) = 1$ for all $n$. So $S(a) = 0$ almost surely for each $a > 0$.
But, it's not immediately clear to me how you intend to define your limit.
A: The series $S(a)=\sum_{n=1}^\infty f(x_n)$ has countably many terms, and any countable set has measure $0$ while any interval has positive measure, so $S(1) = 0$ and I believe the limit $\lim_{a->0} S(a) = 0$ as well.
A: First note the following equality of events: 
$$
\big \{ \sum\nolimits_{n = 1}^\infty  {f(X_n )}  > 0 \big \}  =  \cup _{n = 1}^\infty  \{ f(X_n ) = 1\}.
$$
Now, for each $n$,
$$
{\rm P}[f(X_n) = 1] = {\rm P}[X_n = 0] = 0.
$$
Hence,
$$
{\rm P}\big[\sum\nolimits_{n = 1}^\infty  {f(X_n )}  > 0\big] = {\rm P}[ \cup _{n = 1}^\infty  \{ f(X_n ) = 1\} ] \le \sum\nolimits_{n = 1}^\infty  {{\rm P}[f(X_n ) = 1]}  = 0.
$$
So, for any $a>0$, ${\rm P}[S(a)>0] = 0$, hence $S(a)=0$ with probability $1$.
