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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