Itô's formula yields an Itô process

Question

In our course on stochastic analysis, we proved the following version of the one-dimensional Itô formula: Let $\{W_t\}_{t\ge 0}$ be a one-dimensional Brownian motion w.r.t. some (right-continuous and complete ("usual conditions")) filtration $\{\mathfrak{F}_t\}_{t\ge 0}$ and let $\{X_t\}_{t\ge 0}$ be an Itô process, i.e. $X_t=X_0+\int_{0}^t K_s ds+\int_0^t H_s dW_s$ where $X_0$ is $\mathfrak{F}_0$-measurable, $K$ and $H$ both are progressively measurable and satisfy the integrability conditions $\int_0^t|K(s)|ds<\infty$ a.s. and $\int_0^t H(s)^2ds<\infty$ a.s. for all $t\ge 0$. Then, if $f:\mathbb{R}\rightarrow\mathbb{R}$ is a $\mathcal{C}^2$-function, we obtain for each $t\ge 0$: $$f(X_t)=f(X_0)+\int_0^t\underbrace{\left[f'(X_s)K_s+\frac{1}{2}f''(X_s)H_s^2\right]}_{(A)}ds+\int_0^t \underbrace{f'(X_s)H_s}_{(B)}dW_s.$$

Now, it was remarked that $\{f(X_t)\}_{t\ge 0}$ is again an Itô process. Why is this the case? How to prove all those integrability and measurability conditions?

EDIT: The relevant integrability conditions are quite obvious to me now as I realized that $X$ is a continuous process. But, I do not see why (A) and (B) are progressively measurable.

Thank you in advance!

Answer 1

Recall two facts on progressive measurability:

1. Let $(Y_t)_{t \geq 0}$ be a progressively measurable (real-valued) stochastic process and $f: \mathbb{R} \to \mathbb{R}$ Borel-measurable. Then $(f(Y_t))_{t \geq 0}$ is progressively measurable.
2. Any adapted stochastic process with continuous sample paths is progressively measurable.

Since $(X_t)_{t \geq 0}$ has continuous sample paths, we get that $(X_t)_{t \geq 0}$ is progressively measurable. Consequently, also $(f'(X_t))_{t \geq 0}$ and $(f''(X_t))_{t \geq 0}$ are progressively measurable. As the product/sum of two measurable functions is again measurable, this implies that (A) and (B) are progressively measurable.