Using the Dominated convergence theorem in a sequence of Indicator functions Let $Z_t\sim WN(0,\sigma^2)$ be a white noise. Consider a $\text{MA}(q)$ process:
\begin{equation}
X_t^q = \sum_{j=0}^{q} \theta_j Z_{t-j}, \quad X_t = \sum_{j=0}^{\infty} \theta_j Z_{t-j}
\end{equation}
where $\sum_{j=0}^{\infty} \theta_j^2 < \infty$. Fix any $t$ and any $x$, I want to show that:
$$\lim_{q \to \infty}P(X_t^q \leq x ) = P(X_t \leq x)$$
For this, I tried the Dominated convergence theorem: Define $f_q = I_{[\,X_q \, \leq \, x\,]}$ and  $f = I_{[\,X \, \leq \, x\,]}$. It's easy to show that:
$$\int f_q\,dP = P(X_t^q \leq x ), \quad \int f\, dP = P(X_t \leq x ) $$
Also, it's easy to show that $|f_q| \leq 1$.
It only remains to show that  the sequence $f_q$ converges pointwise to $f$ and I'm having a little trouble showing this. I think that the solution have to do with this two items questions:

*

*$X^q_t \to X_t$ pointwise? How I can show this?

*The first item implies that $f_q \to f$ pointwise?

Some help, pls!

Fix a time point $t$.

*

*$\sum_{j=0}^{\infty}\theta_{j}^{2}<\infty$ implies $X_{j}^{q}$ and $X_{t}$ are both in $L^{2}$, also $\mathbb{E}[|X_{t}^{q}-X_{t}|^{2}]=\sigma^{2}\sum_{j=q+1}^{\infty}\theta_{j}^{2}\to0$ as $q\to\infty$, therefore $X_{t}^{q}\overset{L^{2}}{\to}X_{t}$, hence $X_{t}^{q}\overset{\mathbb{P}}{\to}X_{t}$.


*Convergence in probability implies convergence in distribution, the desired result holds for all $x$ such that $F(X_{t})$ is continuous.
