To make Edgar's answer complete: There always does exist a product-measurable choice.

More precisely, if $f\colon\mathbb{R}^n\to L_2(\mathbb{R}^d)$ is measurable then there exists a product measurable function $g\colon\mathbb{R}^n\times\mathbb{R}^d\to\mathbb{R}$ such that $g(x,\cdot)=f(x)$ holds for every $x$. If $f$ is integrable then $g(\cdot,y)$ is integrable for almost every $y$, and the expected equality $$\int_{\mathbb R^n}f(x)\,dx(y)=\int_{\mathbb R^n}g(x,y)\,dx$$ holds for almost every $y$.

Moreover, analogous assertions hold for every (strongly Bochner) measurable/integrable function $f\colon S\to X$ where $S$ is a $\sigma$-finite measure space and $X$ is a (possibly vector-valued) ideal space over a $\sigma$-finite measure space $T$, see Section 4.4 in my monograph https://www.springer.com/de/book/9783540631606.

Edit IIRC, the special case $X=L_p(\mathbb{R}^d)$ (IIRC only the technically simpler case $p<\infty$) is already contained in some footnote in Hille-Phillips famous monograph about semigroups.

To make Gerald Edgar's answer complete: There always does exist a product-measurable choice.

More precisely, if $f\colon\mathbb{R}^n\to L_2(\mathbb{R}^d)$ is measurable then there exists a product measurable function $g\colon\mathbb{R}^n\times\mathbb{R}^d\to\mathbb{R}$ such that $g(x,\cdot)=f(x)$ holds for every $x$. If $f$ is integrable then $g(\cdot,y)$ is integrable for almost every $y$, and the expected equality $$\int_{\mathbb R^n}f(x)\,dx(y)=\int_{\mathbb R^n}g(x,y)\,dx$$ holds for almost every $y$.

Moreover, analogous assertions hold for every (strongly Bochner) measurable/integrable function $f\colon S\to X$ where $S$ is a $\sigma$-finite measure space and $X$ is a (possibly vector-valued) ideal space over a $\sigma$-finite measure space $T$, see Section 4.4 in my monograph Ideal Spaces (Springer, Berlin 1997).

Edit IIRC, the special case $X=L_p(\mathbb{R}^d)$ (IIRC only the technically simpler case $p<\infty$) is already contained in some footnote in Hille-Phillips famous monograph about semigroups.

To make Edgar's answer complete: There always does exist a product-measurable choice.

More precisely, if $f\colon\mathbb{R}^n\to L_2(\mathbb{R}^d)$ is measurable then there exists a product measurable function $g\colon\mathbb{R}^n\times\mathbb{R}^d\to\mathbb{R}$ such that $g(x,\cdot)=f(x)$ holds for every $x$. If $f$ is integrable then $g(\cdot,y)$ is integrable for almost every $y$, and the expected equality $$\int_{\mathbb R^n}f(x)\,dx(y)=\int_{\mathbb R^n}g(x,y)\,dx$$ holds for almost every $y$.

Moreover, analogous assertions hold for every (strongly Bochner) measurable/integrable function $f\colon S\to X$ where $S$ is a $\sigma$-finite measure space and $X$ is a (possibly vector-valued) ideal space over a $\sigma$-finite measure space $T$, see Section 4.4 in my monograph https://www.springer.com/de/book/9783540631606.

To make Edgar's answer complete: There always does exist a product-measurable choice.

More precisely, if $f\colon\mathbb{R}^n\to L_2(\mathbb{R}^d)$ is measurable then there exists a product measurable function $g\colon\mathbb{R}^n\times\mathbb{R}^d\to\mathbb{R}$ such that $g(x,\cdot)=f(x)$ holds for every $x$. If $f$ is integrable then $g(\cdot,y)$ is integrable for almost every $y$, and the expected equality $$\int_{\mathbb R^n}f(x)\,dx(y)=\int_{\mathbb R^n}g(x,y)\,dx$$ holds for almost every $y$.

Moreover, analogous assertions hold for every (strongly Bochner) measurable/integrable function $f\colon S\to X$ where $S$ is a $\sigma$-finite measure space and $X$ is a (possibly vector-valued) ideal space over a $\sigma$-finite measure space $T$, see Section 4.4 in my monograph https://www.springer.com/de/book/9783540631606.

Edit IIRC, the special case $X=L_p(\mathbb{R}^d)$ (IIRC only the technically simpler case $p<\infty$) is already contained in some footnote in Hille-Phillips famous monograph about semigroups.