By measureable function I mean an "ordinarily" measureable function, that is measureable in a sense of this definition: a function between measurable spaces is said to be measurable if the preimage of each measurable set is measurable.

Let $(X,\ \mathcal{F},\ \mu)$ be a measure space and let $V$ be either $\mathbb{R}^n$ or $\mathbb{C}^n$ with a standard norm $\|v\|=(\sum_{k=1}^n|v_k|^2)^\frac{1}{2}$ for $v=(v_1,\ ...,\ v_n)$ in $V$.

Do the measureable functions (where $V$ is equipped with a $\sigma$-algebra of Borel sets of either $\mathbb{R}^n$ or $\mathbb{C}^n$) form a vector space?

Let $1\leq p \leq \infty$. For a measureable function $f: X\to V$ we define

$$\|f\|_p=(\int\limits_X \|f\|^p\mathrm{d}\mu)^\frac{1}{p}.$$

Do the measureable functions $f$ such that $\|f\|_p<\infty$, after we identify those that are equal almost everywhere, form a Banach space under norm $\|\cdot\|_p$?

I ask, because I know this is not true in the case of functions with values in infinite dimensional Banach spaces. There a notion of Bochner measurable function is introduced.


In the case of finite dimensional spaces, measurability can be reduced to the "ordinary" notion of measurability as follows:

A function $f : X \to \Bbb{R}^n$ is measurable if and only if for each $i \in \{1, \dots, n\}$, the component function $f_i : X \to \Bbb{R}$ is measurable.

This can be seen as follows: Measurability is preserved by continuous functions (because these are Borel-measurable and compositions of measurable functions are measurable). Hence, if $f$ is measurable, then each $f_i = \pi_i \circ f$ is also measurable, because $\pi_i : \Bbb{R}^n \to \Bbb{R}, x \mapsto x_i$ is continuous.

On the other hand, each of the maps $\iota_i : \Bbb{R} \to \Bbb{R}^n, x \mapsto (0, \dots, 0, x, 0,\dots,0)$ is continuous (where the $x$-entry is in the $i$-th component). Hence, if each $f_i$ is measurable, then

$$ f = \sum_{i=1}^n \iota_i \circ f_i $$

is also measurable.

The same arguments remain valid in the case of $\Bbb{C}^n$. In that case, one can even reduce further to measurability of $\mathrm{Re}(f_i)$ and $\mathrm{Im}(f_i)$ for each $i$.

Regarding completeness, one can easily see the norm equivalence

$$ \Vert f \Vert_p \asymp \sum_{i=1}^n \Vert f_i \Vert_p, $$

which allows to reduce the completeness to the $1$-dimensional setting.

Regarding the Bochner measurability, one can show (cf. Serge Lang, Real and Functional Analysis) that $f : X \to E$ (with a Banach space $E$) is $\mu$- Bochner measurable if and only if

  1. There is a $\mu$-null-set $N \subset X$ such that $f(X\setminus N)$ is separable and so that $f|_{X \setminus N}$ is (Borel) measurable in the sense described by you.
  2. $f$ vanishes outside of a set of $\sigma$-finite measure.

Here, a function $f$ is called $\mu$-Bochner-measurable, if there is a sequence of simple functions $f_n = \sum_{i=1}^{m_n} \alpha_i^{(n)} \chi_{M_i^{(n)}}$ with each $M_i^{(n)}$ of finite measure and with $f_n(x) \to f(x)$ for $\mu$-almost all $x$.

This shows that as long as $E$ is separable (this is in particular the case if $E$ is finite dimensional) and $\mu$ is $\sigma$-finite (and complete), there is no difference between ordinary measurability and Bochner measurability. Problems occur as soon as the target space $E$ is not separable.

  • $\begingroup$ Thank you for this very helpful answer. $\endgroup$ – Tom Sep 2 '14 at 23:44

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