I am trying to compute the stochastic integral $$\int_{(0,t]}\mathbb{1}_{[a,b)}(s)dM_s$$ where $0 < a < b< \infty$ are constant and $M$ is a continuous local $L^2$-martingale. I am guessing the answer should be something like $M_{t\wedge b} - M_{t\wedge a}$ but I have to prove that so here is my attempt:
First of all, $\mathbb{1}_{[a,b)}$ is predictable as it can be approximated by the sequence of left-continuous functions $\Pi_{m=1}^n\mathbb{1}_{(a-\frac{1}{m},b-\frac{1}{m}]}$. Furthermore, for any stopping time $\sigma$, $\mathbb{1}_{[0,\sigma]}\mathbb{1}_{[a,b)} \in \mathcal{L}^2(M)$ for any $L^2$-martingale $M$. So $\mathbb{1}_{[a,b)}(s)$ is valid as an integrand.
Now, let $(\tau_k)$ be a localizing sequence for the pair $(\mathbb{1}_{[a,b)}, M)$. The definition of the stochastic integral is that $$\int_{(0,t]}\mathbb{1}_{[a,b)}(s)dM_s = \int_{(0,t]}\mathbb{1}_{[0,\tau_k]}\mathbb{1}_{[a,b)}(s)dM_s^{\tau_k}$$ for any $k$ such that $t \leq \tau_k$. Here the right hand side is the "usual" stochastic integral with respect to the $L^2$-martingale $M^{\tau_k}$. Since I am allowed to choose $k$ as big as I want, I restrict myself to $k$ such that $\tau_k > b$. This gives $$ \int_{(0,t]}\mathbb{1}_{[0,\tau_k]}\mathbb{1}_{[a,b)}(s)dM_s^{\tau_k} = \int_{(0,t]}\mathbb{1}_{[a,b)}(s)dM_s^{\tau_k}$$
As much as I hate technicalities I am not going to jump the gun here and compute the right hand side as rigorously as possible and I require some assistance here.
First I will show that the process $\mathbb{1}_{(a-\frac{1}{n},b-\frac{1}{n}]}(s)$ approximates $\mathbb{1}_{[a,b)}(s)$ in $\mathcal{L}^2(M^{\tau_k})$. For that the following must hold for any $T<\infty$. $$E\int_0^T (\mathbb{1}_{(a-\frac{1}{n},b-\frac{1}{n}]}(s)-\mathbb{1}_{[a,b)}(s))^2d[M^{\tau_k}]_s \rightarrow 0$$ as $n\rightarrow\infty$. I can choose $T$ large enough since the integrand is non-negative and $n$ large enough since I will let $n$ go to infinity later anyway. The above limit then reduces to $$E\left[[M^{\tau_k}]_b-[M^{\tau_k}]_{b-\frac{1}{n}}\right] + E\left[[M^{\tau_k}]_a-[M^{\tau_k}]_{a-\frac{1}{n}}\right] \rightarrow 0$$ For a continuous local $L^2$-martingale $N$ and stopping time $\sigma$, $[N^\sigma] = [N]^\sigma$. Applying this to the expression above, combined with the fact that $b < \tau_k$, I get $$E\left[[M]_b-[M]_{b-\frac{1}{n}}\right] + E\left[[M]_a-[M]_{a-\frac{1}{n}}\right] \rightarrow 0$$ I know that $[M]$ is continuous thanks to the continuity of $M$. So $$[M]_b-[M]_{b-\frac{1}{n}}\rightarrow 0 \quad \text{pointwise}$$ Here is where my first question pops. How do I get to $$E\left[[M]_b-[M]_{b-\frac{1}{n}}\right]\rightarrow 0$$ from this? This seems like a perfect place to apply dominated convergence but I don't see how I could bound $\lvert[M]_b-[M]_{b-\frac{1}{n}}\rvert$. There is one thing I could do and that is the following. Since $M^{\tau_k}$ is an $L^2$-martingale, $E[[M^{\tau_k}]_b] = E[(M^{\tau_k}_b)^2 - (M^{\tau_k}_0)^2] = E[(M_b)^2 - (M_0)^2]$. Then this part of the problem will be solved when I have $$E\left[(M_b)^2-(M_{b-\frac{1}{n}})^2\right]\rightarrow 0$$
Assuming that the limit above is true, I compute \begin{align}\int_{(0,t]}\mathbb{1}_{(a-\frac{1}{n},b-\frac{1}{n}]}(s)\ dM^{\tau_k}_s =& M^{\tau_k}_{b-\frac{1}{n}\wedge t}-M^{\tau_k}_{a-\frac{1}{n}\wedge t}\\ =& M_{b-\frac{1}{n}\wedge t}-M_{a-\frac{1}{n}\wedge t} \end{align} The second equality follows from $t \leq \tau_k$.
So now I want to show that $$M_{b-\frac{1}{n}\wedge t}-M_{a-\frac{1}{n}\wedge t} \rightarrow M_{b\wedge t}-M_{a\wedge t} \quad \text{in} \quad \mathcal{M}^2$$ It is clear that the convergence occurs pointwise but for convergence in $\mathcal{M}^2$ the following must hold for all $t < \infty$, $$E\left[\left(M_{b-\frac{1}{n}\wedge t}-M_{a-\frac{1}{n}\wedge t}-M_{b\wedge t}+M_{a\wedge t}\right)^2\right]\rightarrow 0$$ It seems like dominated convergence would come to the rescue here as well but then again I don't see how to bound $\left(M_{b-\frac{1}{n}\wedge t}-M_{a-\frac{1}{n}\wedge t}-M_{b\wedge t}+M_{a\wedge t}\right)^2$. So this was my second question and the third one is about whether what I wrote (my approach,that is) is mathematically sound.
Question 1: Note that $[M]^{\tau_k}_{t \geq 0}$ is an increasing process. Therefore, $$|[M]^{\tau_k}_b - [M]^{\tau_k}_{b-\frac{1}{n}}| \leq 2 [M]_b.$$ Since $$\mathbb{E}([M]_b^{\tau_k}) = \mathbb{E}((M_b^{\tau_k})^2)<\infty,$$ we can apply the dominated convergence theorem to conclude $$\mathbb{E}([M]_b^{\tau_k}-[M]_{b-1/n}^{\tau_k}) \to 0 \qquad \text{as} \, \, (n \to \infty).$$ (If you are wondering why there is still the "$\tau_k$" in there, then have a look at question 3.)
Question 2: Recall that, by Itô's isometry, $$\mathbb{E}\left( \left| \int_0^{t} g(s) \, dM_s^{\tau_k} \right|^2 \right) = \mathbb{E} \left( \int_0^t g(s)^2 \, d[M]_s^{\tau_k} \right)$$ whenever the expression at the right-hand side is well-defined and $g$ is predictable. Choosing
$$g(s) := 1_{(b-1/n \wedge t, b]}(s) + 1_{(a-1/n \wedge t,a]}(s)$$
we get
$$\begin{align*} \mathbb{E}((M_{b-1/n \wedge t}^{\tau_k} - M_{a-1/n \wedge t}^{\tau_k}-M_{b \wedge t}^{\tau_k}+M_{a \wedge t}^{\tau_k})^2) &= \mathbb{E} \left(\left| \int_0^t g(s) \, dM_{s}^{\tau_k} \right|^2 \right) \\ &= \mathbb{E} \left( \int_0^{t \wedge \tau_k} g(s) \, d[M]_s^{\tau_k} \right) \\ &\stackrel{Q1}{\to} 0. \end{align*}$$
However, we don't even have to do this. We have already shown that the integral $\int_0^{t} 1_{(a,b]}(s) \, dM_s^{\tau_k}$ exists and that $f_n := 1_{(a-1/n,b-1/n]}$ is an approximating sequence of $f:= 1_{(a,b]}$. Therefore, $\int_0^t f_n \, dM_s^{\tau_k}$ converges in $\mathcal{M}^2$ to $\int_0^t f \, dM_s^{\tau_k}$. On the other hand, since $\mathcal{M}^2$-convergence implies almost sure convergence of a subsequence and $$\lim_{n \to \infty} \int_0^t f_n \, dM_s^{\tau_k} = M_{b \wedge t}^{\tau_k}-M_{a \wedge t}^{\tau_k} \qquad \text{a.s.,}$$ it follows from the uniqueness of the limit that $$\int_0^t f \, dM_s^{\tau_k} = M_{b \wedge t}^{\tau_k}-M_{a \wedge t}^{\tau_k}.$$
Question 3: In your answer you write "combined with the that $b<\tau_k$ [...]". Unfortunately, this doesn't work; in general, we cannot choose $k$ sufficiently large such that $b<\tau_k(\omega)$ for all $\omega \in \Omega$. By the definition of a localizing sequence, we know that $\tau_k(\omega) \uparrow \infty$ for all $\omega \in \Omega$, but this does not imply that we can choose $k$ in such a way as you claim it. This means that, during all these technical stuff, we have to keep the $\tau_k$. However, at the end, if we consider the integral
$$ \left( \int_{(0,t]} 1_{[a,b)}(s) \, dM_s \right)(\omega)$$
for some fixed $\omega \in \Omega$, then we can choose $k$ sufficiently large such that $t < \tau_k(\omega)$ and we obtain
$$ \left( \int_{(0,t]} 1_{[a,b)}(s) \, dM_s \right)(\omega) = M_{t \wedge b}-M_{t \wedge a}.$$