There is a well-established way to extend the familiar definition of the Lebesgue spaces $L^p[0,1]$ (for 1 ≤ p ≤ ∞) to the setting of E-valued functions on $[0,1]$, where E is a complex-valued Banach space. I say “well-established” but it was only relatively recently (c. 2013/2014) that I found myself needing to look up some of the technical subtleties, and even more recently I realised I had remained ignorant of some fairly important foundational aspects of this process or construction.

When 1≤ p < ∞ the natural guess is that we should take equivalence classes of those “measurable” functions $f: [0,1]\to E$ that satisfy $(1)\qquad\qquad \displaystyle \int_0^1 \Vert f(t) \Vert^p \,dt < \infty$

But we run into an immediate issue: which functions $[0,1]\to E$ do we declare to be "measurable"? It is clearly necessary for $t\mapsto \Vert f(t)\Vert$ to be measurable as a function $[0,1]\to \mathbb R$ (as well as p-integrable), but is this a good definition or should it be treated as a consequence of some “better” definition?

Moreover: how should we treat the case p=∞ ?

## Digression on Borel versus Lebesgue

We shall work with the Borel sigma-algebra $\mathcal B$ on [0,1] rather than the Lebesgue sigma-algebra $\mathcal L$, though this is largely a matter of preference/convenience: see this EOM entry for some of the relevant context. In the sense of these lecture notes by J. van Neerven, we are working with (strongly or weakly) $\mathcal B$-measurable functions, rather than (strongly or weakly) μ-measurable functions where μ is taken to be Lebesgue measure.

## The key definitions

Definition 1. A function $f:[0,1]\to E$ is called a simple function if there are (Borel) measurable subsets $X_1,\dotsm X_n$ and $\xi_1, \dots, \xi_n$ in $E$ such that $(2)\qquad \qquad \displaystyle f = \sum_{j=1}^n \xi_j 1_{X_j}$

Definition 2. A function $f:[0,1]\to E$ is said to be strongly measurable if there is a sequence $(f_n)$ of simple E-valued functions such that $\Vert f(t) - f_n(t) \Vert \to 0$ for all $t\in [0,1]$.

Definition 3. A function $f:[0,1]\to E$ is said to be weakly measurable if, for each $\psi\in E^*$, the scalar-valued function $\psi\circ f:[0,1]\to \mathbb C$ is (Borel-to-Borel) measurable.

Definition 4. A function $f:[0,1]\to E$ is said to be Borel measurable if $f^{-1}(S)$ is Borel for every Borel subset $S\subseteq E$.

It is a straightforward exercise to show that if $(f_n)$ is a sequence of Borel measurable functions $[0,1]\to\mathbb R$ and $f_n \to f$ pointwise, then $f$ is Borel measurable. See e.g. this MO answer. The same MO answer also explains why strongly measurable E-valued functions are both weakly measurable and Borel measurable. (Essentially, one verifies this for simple functions, and then invokes the result mentioned above.)

The image of any strongly measurable function is always separable (with respect to the norm topology of E), and so if $f:[0,1]\to E$ is strongly measurable, it is weakly measurable and takes values in a separable subspace of E. An important theorem of Pettis tells us that the converse is true.

Theorem 5 (Pettis measurability theorem). Let $f:[0,1]\to E$ be a weakly measurable function which takes values in a separable subspace of E. Then $f$ is strongly measurable.

(One can weaken the hypotheses slightly: it is enough to require separable range and "σ(E,F)-measurability", where F is a weak-star dense subspace of E*. See the notes of van Neerven that are linked above.)

Note that if $\phi: E\to F$ is any function between two Banach spaces, not necessarily linear, and $f:[0,1]\to E$ is a simple function, then $\phi\circ f : [0,1]\to F$ is also a simple function. With this in mind, the following result is an almost immediate consequence of Definition 1.

Proposition 6. Let E and F be Banach spaces and let $\phi:E \to F$ be a continuous function. If $f:[0,1]\to E$ is strongly measurable, then so is $\phi\circ f:[0,1]\to F$.

In particular, if $f:[0,1]\to E$ is strongly measurable, then $t\mapsto \Vert f(t)\Vert$ is (Borel-to-Borel) measurable.

Remark 7. Proposition 6 may seem unsurprising, given that the corresponding statement where “strongly” is replaced by “Borel” is extremely easy to prove. However, it’s not clear to me right now if the result still holds when “strongly” is replaced by “weakly”.

## Lebesgue-Bochner spaces for finite p

In view of the comment following Proposition 6, we can now define E-valued $L^1[0,1]$ in a natural way.

Definition 8. Let E be a Banach space. We say that a strongly measurable function $f:[0,1]\to E$ is Bochner integrable if $\int_0^1 \Vert f(t)\Vert \,dt < \infty$. The vector space V of all such $f$ carries an obvious/natural seminorm, and we define $L^1([0,1];E)$ to be the quotient of V by the kernel of this seminorm.

We view elements of $L^1([0,1];E)$ to be equivalence classes of Bochner integrable E-valued functions, with the equivalence relation being “differs only on a set of measure zero”.

Loosely speaking, for 1≤p< ∞, we can then define $L^p([0,1];E)$ to be the space of all (equivalence classes of) strongly measurable functions $f:[0,1]\to E$ for which $t\mapsto \Vert f(t)\Vert^p$ is integrable. Equipping $L^[([0,1];E)$ with the natural norm, one can show it is complete by adapting the usual proofs from the scalar-valued case.

It is not hard to prove that the simple functions $[0,1]\to E$ form a norm-dense subspace of $L^p([0,1];E)$, and so if one likes tensor product formalism one can view $L^p([0,1];E)$ as a completion of the algebraic tensor product $L^p[0,1] \otimes E$.