Suppose $(\mathcal{F}_t)_{t \geq0}$ is a filtration on a probability space and $W=(W_t)_{t \geq0}$ is a Brownian Motion with respect to this filtration. Let $(X_t)_{t \geq0}$ denote some stochastically integrable process. I want to prove the following for real numbers $a<b$: $$ E\bigg[\bigg(\int_a^b X_s dW_s \bigg)^2 \ \bigg\vert \ \mathcal{F}_a \bigg] =\int_a^b (X_s)^2 ds $$ where $ds$ denotes the Lebesgue measure.
I know that $\int_0^b X_s dW_s $ is a again a martingale at time point $b$ as $W$ is a martingale. We can write $$ \int_a^b X_s dW_s = \int_0^b X_s dW_s - \int_0^a X_s dW_s $$ and then write out the left hand side out as a square and somehow use the martingale property I suspect. But I am still unsure how it turns into a Lebesgue integral and that the square ends up inside... Any help is appreciated!