Weak convergence and Fourier transform For positive integer $k$, let $\mu_k=\dfrac{1}{2}\left(\delta(x)+\delta\left(x-\dfrac{2}{3^k}\right)\right)$. Let $dC_k=\mu_1\ast\cdots\ast\mu_k$. We have that $dC_k$ converges weakly to $\mu_C$, where $C$ is the Cantor function.
Now we want to show that $\hat{\mu}_C(y)=e^{ay}\prod_{k=1}^\infty\cos(y/3^k)$ for some $a\in\mathbb{C}$. For this, we also need to show that the infinite product converges.
I want to use the weak convergence. Is it true that $\hat{dC_k}(y)$ converges to $\hat{\mu}_C(y)$?
 A: In my answer to your previous question, I showed that $dC_k \rightharpoonup dC$, where $dC$ is the Cantor measure.  One way to proceed from here to the convergence of the Fourier transforms of these measures is to try to realize the entire $dC_k$ sequence and $dC$ on the same measure space so that we can apply dominated convergence directly.  
If you look at the argument I gave in the previous thread, we implicitly did this.  If you take as your measure space the natural one for the coin flip sequence $X_j$, then the law of $\sum_{j=1}^k \frac{X_j}{3^j}$ is $dC_k$ and the law of $\sum_{j=1}^\infty \frac{X_j}{3^j}$ is $dC$.  This means that if we call $P$ the law of the coin flip sequence, then 
\begin{align*}
E\left[e^{it \sum_{j=0}^k \frac{X_j}{3^j}} \right] &= \int e^{it \sum_{j=0}^k \frac{X_j(\omega)}{3^j}} dP(\omega) \\
&= \int e^{itx} dC_k(x) \\
&= \widehat{dC}_k(t),
\end{align*}
where the first equality is by definition and the second equality comes from the argument I gave in my previous answer.  Similarly, 
\begin{align*}
E\left[e^{it \sum_{j=0}^\infty \frac{X_j}{3^j}} \right] &= \int e^{it \sum_{j=0}^\infty \frac{X_j(\omega)}{3^j}} dP(\omega) \\
&= \int e^{itx} dC(x) \\
&= \widehat{dC}(t).
\end{align*}
$dP$ is a probability measure (remember, it is the law of a sequence of $\{0,2\}$ valued fair coin flips), so $1$ is integrable. I showed in my last answer that $\sum_{j=0}^k \frac{X_j(\omega)}{3^j} \to \sum_{j=0}^\infty \frac{X_j(\omega)}{3^j}$.  We can therefore apply dominated convergence to the functions $e^{it \sum_{j=0}^k \frac{X_j(\omega)}{3^j}}$ to get the desired limit.
If you are not comfortable with that approach, here is another option.  $e^{itx}$ is a bounded continuous function and by the Portmanteau theorem, a sequence of probability measures converges weakly if and only if for all bounded continuous functions the expectations converge.  Again, we already showed weak convergence in the previous question.  It does matter here that our measures are all probability measures, by the way.  That result fails for general measures and general weak convergence.
All that is left is to compute the characteristic functions.  Notice that 
\begin{align*}
\widehat{\mu}_j(t) &= \frac{1}{2}\left(1 + e^{i\frac{2t}{3^j}}\right) \\
&= e^{i\frac{t}{3^j}}\underbrace{\frac{1}{2}\left(e^{-i\frac{t}{3^j}} + e^{i\frac{t}{3^j}}\right)}_\textrm{$\cos \left( \frac{t}{3^j}\right)$}
\end{align*}
so that
\begin{align*}
\widehat{dC}_k (t) = e^{i\sum_{j=1}^k \frac{t}{3^j}} \prod_{j=1}^k \cos\left(\frac{t}{3^j} \right).
\end{align*}
Taking limits (and the limits exist by what we did previously), we find that
\begin{align*}
\widehat{dC}(t) &= e^{i\sum_{j=1}^\infty \frac{t}{3^j}} \prod_{j=1}^\infty \cos\left(\frac{t}{3^j} \right) \\
 &= e^{i\frac{t}{2}} \prod_{j=1}^\infty \cos\left(\frac{t}{3^j} \right).
\end{align*}
