I have a question regarding the quadratic variation of a martingale. Assume for each $x \in \mathbb{R}$ there exists a martingale $(M_t(x))_{t \geq 0}$ with known quadratic variation $\langle M(x) \rangle _t$. Assume $\phi : \mathbb{R} \to \mathbb{R}$ is a sufficiently smooth function and rapidly decreasing at $\infty$, such that $\int_{\mathbb{R}}f(x)\phi(x)\mathrm{d}x$ is finite and well-defined for all 'nicely behaved' continuous real valued functions $f$ (I skip some details here, the main part is that one doesn't have to worry about convergence of integrals in that question).
First, I want to verify $(\int_{\mathbb{R}}M_t(x)\phi(x)\mathrm{d}x)_{t \geq 0}$ is a martingale. Is it true under the above assumptions? I am not sure how to formalize this, so any help will be greatly appreciated. I belive it requires using the Fubini's theorem, but I am not sure how to justify its use here.
Second, after verifying the above I want to calculate the quadratic variation of the above martingale. That is, I want to calculate $\langle\int_{\mathbb{R}}M_{(\cdot)}(x)\phi(x)\mathrm{d}x \rangle _t$ in terms of $\langle M(x) \rangle _t$ and $\phi$. Here I really have know idea what to do. Again, any suggestion will be great.
For the first question, yes, this is a martingale. This only requires that $\int \mathbb{E}[|M_t(x)|] |\phi(x)| dx < \infty$ for all $t$. Then, by Fubini's theorem, \begin{align*} \mathbb{E}\left[ \int M_t(x) \phi(x) dx \mid \mathcal F_s \right] &= \int \mathbb{E}[M_t(x)|\mathcal F_s] \phi(x) dx \\ &= \int M_s(x) \phi(x) dx . \end{align*} If you're a little skeptical of using Fubini's theorem with conditional expectation, I'll put a proof that it's valid at the end.
Now we will compute the quadratic variation. For notational convenience, let $X_t := \int M_t(x) \phi(x) dx$. By the definition of quadratic variation, \begin{align*} \langle X,X \rangle_t &= \lim_{n \rightarrow \infty} \sum_{k=0}^n (X_{t_{k+1}}-X_{t_k})^2 \end{align*} For a given partition $(t_0,...,t_n)$ of $[0,t]$, we have \begin{align*} \sum_{k=0}^n (X_{t_{k+1}}-X_{t_k})^2 &= \sum_{k=0}^n \left( \int M_{t_{k+1}}(x) \phi(x) dx - \int M_{t_k}(x) \phi(x) dx \right)^2 \\ &= \sum_{k=0}^n \left( \int \big(M_{t_{k+1}}(x) - M_{t_k}(x)\big) \phi(x) dx \right)^2 \\ &= \sum_{k=0}^n \left( \int \big(M_{t_{k+1}}(x) - M_{t_k}(x)\big) \phi(x) dx \right)\left( \int \big(M_{t_{k+1}}(y) - M_{t_k}(y)\big) \phi(y) dy \right) \\ &= \sum_{k=0}^n \left( \int \big(M_{t_{k+1}}(x) - M_{t_k}(x)\big) \phi(x) dx \right)\left( \int \big(M_{t_{k+1}}(y) - M_{t_k}(y)\big) \phi(y) dy \right) \\ &= \sum_{k=0}^n \int \int \big(M_{t_{k+1}}(x) - M_{t_k}(x)\big) \big(M_{t_{k+1}}(y) - M_{t_k}(y)\big) \phi(y) \phi(x) dx dy \\ &= \int \int \sum_{k=0}^n \big(M_{t_{k+1}}(x) - M_{t_k}(x)\big) \big(M_{t_{k+1}}(y) - M_{t_k}(y)\big) \phi(y) \phi(x) dx dy. \end{align*}
Now, assuming everything is well-behaved enough that we can exchange the limits with the integrals, we have
\begin{align*} &\lim_{n \rightarrow \infty}\int \int \sum_{k=0}^n \big(M_{t_{k+1}}(x) - M_{t_k}(x)\big) \big(M_{t_{k+1}}(y) - M_{t_k}(y)\big) \phi(y) \phi(x) dx dy \\ &= \int \int \lim_{n \rightarrow \infty} \sum_{k=0}^n \left( \big(M_{t_{k+1}}(x) - M_{t_k}(x)\big) \big(M_{t_{k+1}}(y) - M_{t_k}(y)\big) \right)\phi(y) \phi(x) dx dy \\ &= \int \int \langle M(x),M(y) \rangle_t \phi(y) \phi(x) dx dy, \end{align*} so \begin{align*} \langle X,X \rangle_t &= \int \int \langle M(x),M(y) \rangle_t \phi(y) \phi(x) dx dy. \end{align*} The quadratic covariation of $\int M_t(x) \phi(x)dx$ unfortunately requires the cross-variation $\langle M(x),M(y) \rangle_t$ rather than just the quadratic variation.
Proof of conditional Fubini's theorem: Let $A \in \mathcal F_s$. Then, by Fubini's theorem, \begin{align*} \mathbb{E}\left[\int M_t(x) \phi(x) dx 1_A \right] &= \int \mathbb{E}[ M_t(x)1_A] \phi(x) dx \\ &= \int \mathbb{E}[ \mathbb{E}[M_t(x)|\mathcal F_s]1_A] \phi(x) dx \\ &= \mathbb{E}\left[\int \mathbb{E}[M_t(x)|\mathcal F_s] \phi(x) dx 1_A \right]. \end{align*} Since this holds for arbitrary $A \in \mathcal F_s$, we conclude \begin{align*} \mathbb{E}\left[\int M_t(x) \phi(x) dx \mid \mathcal F_s \right] &= \int \mathbb{E}[M_t(x)|\mathcal F_s] \phi(x) dx. \end{align*}