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Let $f: \mathbb{R}^n \to L^2(\mathbb{R}^d) $ be a Bochner-integrable function (all measures are the Lebesgue measure). Does then $ \int_{\mathbb{R}^n} f(x) d\lambda^n (y) = \int_{\mathbb{R}^n} f(x)(y) d\lambda^n $ hold for $\lambda^d$-almost all $y \in \mathbb{R}^d$? I.e. can one compute such Bochner integrals just by computing ordinary Lebesgue integrals?

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Answer: YES and NO.

YES: In any practical situation you are likely to meet, your formula is correct. You would prove it using Fubini's Theorem, pairing your two sides with an arbitrary $h \in L^2(\mathbb R^d)$ and getting the same answer on both sides. The catch is, you have to be able to apply Fubini.

NO: As stated, it can fail. $f(x) \in L^2(\mathbb R^d)$, so $f(x)$ is an equivalence class. For each $x$, CHOOSE some representative for that class, call it $f(x)(y)$. But now, for fixed $y$ it may fail that $f(x)(y)$ is a measurable function of $x$. Or even if those are all measurable, it may fail that $f(x)(y)$ is measurable in the product measure.

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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.

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  • $\begingroup$ May I ask if such measurable representation exists if $X=L_\infty (\mathbb{R}^d)$? $\endgroup$
    – Akira
    Aug 10, 2023 at 13:36

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