# Limit of convolution of measures is Cantor function

For positive integer $k$, let $\mu_k=\dfrac{1}{2}\left(\delta(x)+\delta\left(x-\dfrac{2}{3^k}\right)\right)$. Show that $$\lim_{k\rightarrow\infty}(\mu_1\ast\mu_2\ast\cdots\ast\mu_k)((-\infty,x))=C(x),$$ where $C$ is the Cantor function.

The definition of measure convolution that I find is

$$(\mu_1\ast \mu_2)(I)=\int_\mathbb{R}\int_\mathbb{R}1_I(x_1+x_2)d\mu_1(x_1)d\mu_2(x_2)$$ for any interval $I$.

I'm not sure how to start on this. It's a convolution of $k$ measures...

Here is a probabilistic proof. Recall that a convolution of measures corresponds to taking a sum of independent random variables. Recall that $\mu_k$ is the law of a random variable which takes on value $0$ with probability $\frac{1}{2}$ and $\frac{2}{3^k}$ with probability $\frac{1}{2}$. Therefore, if $X_k$ is sequence of $iid$ random variables taking on the value $2$ with probability $\frac{1}{2}$ and $0$ with probability $\frac{1}{2}$, $\mu_k$ is the law of $\frac{X_k}{3^k}$.
$\mu_1 * \dots * \mu_k$ is therefore the law of $\sum_{i=1}^k \frac{X_i}{3^i}$. It is not hard to see that this sum converges absolutely almost surely (and therefore in distribution) to $\sum_{i=1}^\infty \frac{X_i}{3^i}$ and that the limit random variable is uniformly distributed on the set of numbers which have only $0$ or $2$ in their tenary expansion. This characterizes the limit as the uniform measure on the Cantor set. The Cantor function is the distribution function of that measure.