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For a Lebesgue integral, I generally see two equivalent definitions of the integral of a function $f:X\rightarrow\mathbb{R}$. One is based on Lebesgue sum $$ \lim_{n\rightarrow\infty}\sum_{k\in\mathbb{Z}}\frac{k}{2^n}\mu\left(\left\{x\in X : f(x)\in\left[\frac{k}{2^n},\frac{k+1}{2^n}\right]\right\}\right) $$ the other is based on simple functions. Now, for a Bochner integral, I only see definitions based on simple functions. Is there an equivalent definition that looks similar to a Lebesgue sum?

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Per Michael's comment, for a function $f: X\rightarrow \mathbb{R}^d$, we may be able to define something like

$$ \lim_{n\rightarrow\infty}\sum_{k_1\in\mathbb{Z}}\dots\sum_{k_d\in\mathbb{Z}}\frac{1}{2^n}\begin{bmatrix}k_1\\\vdots\\k_d\end{bmatrix}\mu\left(\left\{x\in X : f(x)\in\prod_{j=1}^d\left[\frac{k_d}{2^n},\frac{k_d+1}{2^n}\right]\right\}\right), $$ which only works in $\mathbb{R}^d$. Though, at this point, I'm not sure about the requirements on $f$ for this to make sense and whether or not the codomain needs to remain finite dimensional.

wyer33
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I'm pretty sure the answer is no. As I've had it explained to me, the Bochner integral is an extension of the usual Lebesgue integral to general Banach spaces. One of the issues of working in a general Banach space is that you can suddenly have an infinite basis (even worse - uncountably infinite: sauce Let $X$ be an infinite dimensional Banach space. Prove that every Hamel basis of X is uncountable.). In particular, we wouldn't be able to write a product of intervals inside a measure like we can for Lebesgue integrable functions. As a result, all Lebesgue integrable functions are Bochner integrable, but not the other way around.

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cnick
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