This is from Shreve's Stochastic Calculus for Finance II, Appendix A.3. Consider infinitely many independent coin tosses with outcomes $\omega_n$, $n = 1, 2, \dots$ such that $\omega_n$ can be either $H$ = heads or $T$ = tails. Define
\begin{equation}
Y_n\left(\omega\right) = \begin{cases}1 & \text{ if }\omega_n = H \\
0& \text{ if } \omega_n = T\text{.}
\end{cases}
\end{equation}
Define $\displaystyle Y = \sum\limits_{n=1}^{\infty}\dfrac{2Y_n}{3^{n}}$. Shreve shows that $Y$ can only take values in the Cantor set, and since that particular set has measure zero, its Lebesgue measure is zero as well, thus it does not have a probability density function (since integrating such a density function would give a value of $0$). This part I understand, as well as my professor.
Now Shreve claims that $Y$ does not have a probability mass function.
If it did, then for some number $x \in C$ [the Cantor set] we would have $\mathbb{P}\left(Y=x\right) > 0$. But $x$ has a unique base-three expansion
\begin{equation}
x = \sum\limits_{n=1}^{\infty}\dfrac{x_n}{3^{n}}\text{,}
\end{equation}
where each $x_n$ is either 0, 1, or 2 unless $x$ is of the form $\dfrac{k}{3^{n}}$ for some positive integers $k$ and $n$. In the latter case, $x$ has two base-three expansions… In either case, there at most two choices of $w \in \Omega_{\infty}$ [that is, the infinite coin-toss space] for which $Y(\omega) = x$. In other words, the set $\{\omega \in \Omega; Y(\omega) = x\}$ has either one or two elements. The probability of a set with one element is zero, and the probability of a set with two elements is $0 + 0 = 0$. Hence $\mathbb{P}\left\{Y = x\right\} = 0$.
My professor and I were able to follow this until the statement "The probability of a set with one element is zero." Is this a true statement? How so?
I thought it would be perhaps the fact that the Lebesgue measure of $[a, a] = \{a\}$ is zero, but I don't see any reason why Shreve would only use the Lebesgue measure, since it is not a general probability measure.
Best Answer
For every one element set $A = \{\hat\omega\}$ and every $n$, we have $$\mu(A) = \mu(\{\hat\omega\}) \leq \mu(\{\omega:\omega_1=\hat\omega_1, \dots, \omega_n = \hat\omega_n\}),$$ since $A = \{\hat\omega\} \subset \{\omega:\omega_1=\hat\omega_1, \dots, \omega_n = \hat\omega_n\}$. Since coin tosses are independent, we have $$\mu(\{\omega:\omega_1=\hat\omega_1, \dots, \omega_n = \hat\omega_n\}) = \prod_{i=1}^n \mu(\{\omega:\omega_i=\hat\omega_i\}) = \prod_{i=1}^n \frac{1}{2} = \frac{1}{2^n}.$$ We get that $\mu(A) \leq 1/2^n$ for every $n$. Therefore, $\mu(A) = 0$. We are done.