# Why bother with the space $\mathcal{L}^1$ for integration when we can abstractly deal with the completion of a semi-normed space

I'm studying the Bochner-Lebesue integral, and while I understand the general construction, I have a few questions about the way it is being presented. Typically, the story goes like this:

We start with a measure space $$(X,\mathcal{A}, \mu)$$, and a Banach space $$E$$ (over $$\Bbb{R}$$ or $$\Bbb{C}$$). Then, we can define the space $$S$$ of simple functions $$X\to E$$, and for such simple functions, we can define an integral $$I(\cdot):= \int_X (\cdot) \, d\mu : S \to E$$ in the usual way. Then, we can define a semi-norm $$\lVert\cdot \rVert_1$$ on $$S$$ by setting $$\lVert \phi \rVert_1 := \int_X|\phi|\, d\mu$$ (this is to be thought of as integration when the Banach space is $$E=\Bbb{R}$$, which is of course well-defined). Thus, we have a semi-normed space $$(S, \lVert \cdot \rVert_1)$$.

At this point, we note that $$S$$ need not be complete, which of course very undesirable for analysis. So, all the presentations I've seen start by defining $$\mathcal{L}^1$$ as the space of functions $$X\to E$$ which are the almost-everywhere pointwise limit of Cauchy sequences in $$S$$. Then, one proves that under these hypotheses, we can extend the integral to a map (pardon the reuse of notation) $$I(\cdot)\equiv \int_X(\cdot)\, d\mu:\mathcal{L}^1 \to E$$, and also extend the seminorm $$\lVert \cdot \rVert_1$$ to $$\mathcal{L}^1$$, such that integration is still a continuous map (with operator norm $$\leq 1$$), and that finally, $$(\mathcal{L}^1, \lVert \cdot \rVert_1)$$ is a complete semi-normed space containing the simple functions $$S$$ as a dense subspace. Therefore, by taking the quotient space of $$\mathcal{L}^1/\{\phi\in \mathcal{L}^1: \, \lVert \phi\rVert_1 = 0\}$$, and calling this $$L^1$$, this becomes a Banach space (because by taking this quotient, the semi-norm induces a norm, which is easily verified to be complete). Finally, it is a simple matter of linear algebra to see that we can "transfer" the integration map in the sense that we get a map $$\tilde{I}:L^1 \to E$$, such that $$I = \tilde{I}\circ \pi$$ ($$\pi$$ being the quotient map $$\mathcal{L}^1 \to L^1$$).

The result is that we have an integration operator $$\tilde{I}$$, defined on a Banach space $$L^1$$, which naturally reduces to what we'd like it to be on simple functions. Now, my question is that why do we bother to introduce the space $$\mathcal{L}^1$$ along the way. My thinking is that just as every metric space has a completion, which is uniquely determined up to isometry, we can do a similar thing for semi-normed spaces, in the form of this theorem:

Theorem

Let $$(S, \lVert \cdot \rVert)$$ be a semi-normed space (over real or complex field), and let $$S_0$$ be the subspace of elements with $$0$$ semi-norm. Then, there exists a completion of $$S$$, i.e a pair $$(V,\gamma)$$, where $$V$$ is a Banach space (over the same field), and $$\gamma:S\to V$$ is a map such that

• $$\gamma$$ is linear
• $$\ker(\gamma) = S_0$$
• $$\text{image}(\gamma)$$ is a dense subspace of $$V$$
• $$\gamma$$ preserves seminorms and norms; i.e for all $$s\in S$$, $$\lVert\gamma(s) \rVert_V = \lVert s \rVert_S$$.

Also, this completion is determined up to isomorphism (i.e if we had another such pair, then we can make a nice commutative diagram and then obtain an isomorphism of Banach spaces simply by extending the relevant maps from the dense subspaces to the whole space).

So, when we have the space of simple functions $$S$$, we could apply this theorem to get the Banach space $$V$$ (which is up to isomorphism the same as $$L^1$$ constructed above), and using similar linear algebra trickery, we can induce an integral $$\tilde{I}$$ on a dense subspace of $$V$$, and then extend by continuity to the whole space.

My questions/concerns:

I realize that by the uniqueness aspect of the completion, both these methods give us the same final outcome: a Banach space, and some type of notion of integral, and of course, the first approach is much more concrete and easier to appreciate on first glance. However, I recently read up about completions of metric (semi-)normed spaces, which is why I thought of the second approach. So I guess my question boils down to: is there anything we gain significantly (besides a bit of concreteness) by realizing $$L^1$$ as a certain quotient space of functions, rather than just thinking of $$L^1$$ as an abstract completion of the space of simple functions?

Is it perhaps because thinking of Banach spaces as being (almost) a space of functions, rather than some abstract construction (like equivalence classes of Cauchy sequences) makes it significantly easier to analyze the space in some sense (hence the term "functional" analyis)? If this is the case, I'd appreciate if you could elaborate on why specifically thinking in terms of function spaces makes the analysis easier/clearer/preferable (I'm not too sure what word I should use here).

• You do not easily get Lebesgue's theorem on dominated convergence in that way. This is one of the most powerful tools in almost all branches of analysis. Commented Aug 6, 2020 at 10:25

Otherwise, why even define real valued $$L^p$$ the way we do? Why not just define it as the completion of compactly supported continuous functions under the $$L^p$$ norm? It is because we might more interested in the elements of the $$L^p$$ spaces than the spaces themselves.
• Alright it does make sense that we are also interested in the actual functions themselves, as opposed to the Banach space $L^p$ by itself as a whole (though I'm sure people are interested in this as well). Thank you for your answer Commented Aug 8, 2020 at 6:33
Indeed you can do that. The cost of doing so is then you don't really know whether a function is integrable unless it is a simple function, so you need to work to provide a better characterization of functions in $$L^1(X;E)$$.