Is there an easy proof that the set of $x \in [0,1]$ whose limit of proportion of 1's in binary expansion of $x$ does not exist has measure zero?

Question

So for given $x \in [0,1]$, if we let $f_n(x)$ be the fraction of 1's occurring in the first $n$ binary digits of the binary expansion for $x$ (where we always assume an infinite trailing string of 0 instead of 1 for certain rationals $x$), then the measure of the set $X_c \subset [0,1]$ for which $\lim_{n \to \infty} f_n(x) = c$ for $x \in X_c$ is 1 if $c = 1/2$, and then obviously $0$ otherwise. However this is a non-trivial fact that the measure of $X_{1/2}$ is 1. A consequence, however, is that the measure of the set $Y$ for which the limit does not exist is also $0$. Is there any easy direct way to prove this consequence, without showing that the measure of $X_{1/2}$ is 1? It seems easier to show that the measure is $0$ for $X_c$ when $c \neq 1/2$.