Integration of progressively measurable process Let $X=\{X_{t},\cal{F}_{t}; 0\leq t<\infty\}$ be a progressively measurable process and $f(t,x):[0,\infty)\times \mathbb{R}^{d}\rightarrow \mathbb{R}$ be a bounded, $\cal{B}([0,\infty))\otimes \cal{B}(\mathbb{R}^{d})$-measurable function. Show that the process $Y_{t} =\int_{0}^{t}f(s,X_{s})ds; t\geq0$ is progressively measurable with respect to $\{\cal{F}_{t}\}$
 A: This question can be reduced to looking at a simpler case:
   $$Y_t(\omega) = \int_0^t \widehat{X}_s(\omega)\, ds \, ,\quad \text{where $\widehat{X}$ is progressively measurable and bounded.} $$
Fix $T > 0$,
$\star$mapping 1 denoted by $g_1$) $\quad$     $(t, \omega)\in \big[ 0, T \big]\times \Omega\longmapsto X_t(\omega)\in\mathbb{R}^d$ is $\mathscr{B}\big([0,T]\big)\bigotimes \mathscr{F}_T\big/\mathscr{B}\big( \mathbb{R}^d \big)$, since $X$ is progressive measurable.
$\star$mapping 2 denoted by $g_2$) $\quad$  $(t, \omega)\in \big[ 0, T \big]\times \Omega\longmapsto \big( t, X_t(\omega) \big)\in\mathbb{R}_+\times\mathbb{R}^d$ is $\mathscr{B}\big([0,T]\big)\bigotimes \mathscr{F}_T\big/\mathscr{B}\big(\mathbb{R}_+\big)\bigotimes\mathscr{B}\big( \mathbb{R}^d\big)$ due to the following easy argument
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Note that $\mathscr{B}\big( \mathbb{R}_+\big)\bigotimes\mathscr{B}\big( \mathbb{R}^d\big)$ is generated by $\mathscr{B}\big( \mathbb{R}_+\big)\times\mathscr{B}\big( \mathbb{R}^d\big)$, let $A\times B\in\mathscr{B}\big(\mathbb{R}_+\big)\times\mathscr{B}\big( \mathbb{R}^d\big)$, then $$ g_2^{-1}\Big( A\times B \Big) = \Big[ \, \big( A \cap [ 0, T] \big) \times \Omega \Big]\cap g_1^{-1}\big( B \big)\in \mathscr{B}\big([0,T]\big)\bigotimes\mathscr{F}_T\,  $$
$\Longrightarrow g_2^{-1}\Big( \mathscr{B}\big(\mathbb{R}_+\big)\bigotimes\mathscr{B}\big( \mathbb{R}^d\big)\Big) \subset \mathscr{B}\big([0,T]\big)\bigotimes\mathscr{F}_T$ " 
$\star$mapping 3 denoted by $g_3 := f\circ g_2 $) $\quad$ 
$(t, \omega)\in \big[ 0, T \big]\times \Omega\longmapsto \big( t, X_t(\omega) \big)\in\big[ 0, T \big]\times\mathbb{R}^d\longmapsto f\big(  t, X_t(\omega) \big)\in\mathbb{R}$, therefore,
 $$g_3^{-1}\Big( \mathscr{B}(\mathbb{R}) \Big) = g_2^{-1}\circ f^{-1}\Big( \mathscr{B}(\mathbb{R}) \Big)\subset g_2^{-1}\Big( \mathscr{B}\big(\mathbb{R}_+\big)\bigotimes\mathscr{B}\big( \mathbb{R}^d\big)\Big) \subset \mathscr{B}\big([0,T]\big)\bigotimes\mathscr{F}_T\, , $$
that is,
$(t, \omega)\in \big[ 0, T \big]\times \Omega\longmapsto  f\big(  t, X_t(\omega) \big)\in\mathbb{R}$ is  $\mathscr{B}\big([0,T]\big)\bigotimes\mathscr{F}_T\big/\mathscr{B}\big( \mathbb{R} \big)$ and bounded.

Here is a quick solution:
Since $\big(\widehat{X_s}, s\geq 0 \big)$ is progressively measurable, then it is adapted and so is $Y$. Due to the boundedness, $Y$ is also continuous. As we know (for example, 1.13 Proposition on Page 5 in Karatzas and Shreve) that continuous+adaptedness imply progressive measurability. We are done. 
Q.E.D.
