# Tensor Product Of $L^p$ Spaces Is Dense In The Product $L^p$ Space

I want to prove the following:

Proposition: Let $$(X,\mathcal{A},\mu)$$ and $$(Y, \mathcal{B}, \nu)$$ be $$\sigma$$-finite measure spaces and let $$(X\times Y , \mathcal{A} \otimes \mathcal{B} , \mathcal{\mu}\otimes \mathcal{\nu})$$ be the product measure space. Further let $$p \in [1, \infty)$$. Then $$\overline{L^p(X) \otimes L^p(Y)} = L^p(X \times Y),$$ where the closure is with respect to the norm in $$L^p(X \times Y)$$.

Here $$L^p(X) \otimes L^p(Y)$$ is identified with a subspace of $$L^p(X \times Y)$$ through the natural embedding $$f \otimes g \mapsto \big( (x,y) \mapsto f(x) g(y) \big)$$.

I have proven the statement in the special case when $$\mu$$ and $$\nu$$ are finite (not $$\sigma$$-finite) and my question is:

Is my proof of the finite measure case correct and if it is, how can i adapt it to the case where the measures are $$\sigma$$-finite?

My proof (of the finite case): It suffices to show that $$\chi_C \in \overline{L^p(X) \otimes L^p(Y)}$$ for every $$C \in \mathcal{A} \otimes \mathcal{B}$$ with $$\mu \otimes \nu (C) < \infty$$. Here $$\chi_C$$ denotes the indicator function of the set $$C$$. This is because the linear combinations of such indicators are dense in $$L^p(X \times Y)$$.

Now assume that $$\mu$$ and $$\nu$$ are finite (and hence also $$\mu \otimes \nu$$), then we can use the following theorem, which is proven here:

Theorem: Let $$(X,\mathcal B,\mu)$$ be a finite measure space. Let $$\mathcal A\subset \mathcal B$$ be an algebra generating $$\cal B$$. Then for all $$B\in\cal B$$ and $$\varepsilon>0$$, we can find $$A\in\cal A$$ such that $$\mu(A\Delta B)<\varepsilon.$$

Here $$\Delta$$ is the symmetric difference: $$A \Delta B = (A \setminus B ) \cup (B\setminus A) = (A \cup B) \setminus (A \cap B)$$.

To use the theorem define $$\mathcal{E}_0 = \{ A \times B : A \in \mathcal{A}, B \in \mathcal{B} \}$$ and $$\mathcal{E} = \big \{ \bigcup_{i=1}^n C_i : C_i \in \mathcal{E}_0 , n \in \mathbb{N} \big \}.$$ Then $$\mathcal{E}$$ is an algebra that generates $$\mathcal{A} \otimes \mathcal{B}$$. Clearly the indicator functions of all elements of $$\mathcal{E}_0$$ are in $$L^p(X) \otimes L^p(Y)$$, because they have product form and finite measure. The indicator function $$\chi_E$$ of every member $$E$$ of $$\mathcal{E}$$ is also in $$L^p(X) \otimes L^p(Y)$$, because $$E$$ can be written as a finite and disjoint union of sets in $$\mathcal{E}_0$$ and then $$\chi_E$$ is just the sum of the indicator functions on these rectangle sets.

Now let $$C \in \mathcal{A} \otimes \mathcal{B}$$. Then by the theorem there exist for every $$\varepsilon >0$$ an $$E \in \mathcal{E}$$ so that $$\mu \otimes \nu (C \Delta E )< \varepsilon$$ and therefore $$\| \chi_C - \chi_E \|= \| \chi_{C \Delta E} \| = \big( \mu \otimes \nu (C \Delta E ) \big)^{1/p} < \varepsilon^{1/p},$$ which concludes the proof.

My idea to extend the proof to the $$\sigma$$-finite case is that the $$\sigma$$-finite case can perhaps be reduced to the finite case by restricting the product measure space in some suitable way to make it finite. But i could not come up with a suitable restriction.

• If your space if sigma-finite, then you can write $X=\bigcup_n E_n, Y=\bigcup_m F_m$ where $(E_n)_n, (F_m)_m$ are increasing and of finite measure. Pick $f\in L^p(X\times Y)$ and set $g_k(x,y)=\chi_{E_k}(x)\chi_{F_k}(y) f(x,y)$. We have $g_k\rightarrow f$ in $L^p(X\times Y)$ by dominated convergence. By the finite case you can approximate $g_k$ by linear combinations of products of indicator functions in $L^p$ and you are good to go. I leave it to you to fill in the details. Commented Jul 18, 2023 at 4:32
• @SeverinSchraven Thanks for the proof suggestion. I managed to prove the result. I really have to remember the construction of the sequence $g_k$. I am sure it can be used in many places to transfer results from finite to $\sigma$-finite cases.
– jd27
Commented Jul 18, 2023 at 8:04
• You are very welcome. Indeed, this trick is pretty neat and allows quite often to pass from the finite to the sigma-finite case. Commented Jul 18, 2023 at 16:29
• I think that the statement "The indicator function $\chi_E$ of every member $E$ of $\mathcal{E}$ is also in $L^p(X) \otimes L^p(Y)$" maybe not correct. Let $X=Y=\mathbb R$. Let $$E = (\{1\} \times \{1\}) \cup ( \{2\} \times \{2\}).$$ Then $$1_E (x, y)= \begin{cases} 1 &\text{if} \quad (x,y) =(1, 1) \, \text{ or } \, (x,y) =(2, 2), \\ 0 &\text{otherwise}. \end{cases}$$ I could not find $f \otimes g \in L^p(X) \otimes L^p(Y)$ such that $f \otimes g =1_E$. Commented Aug 17, 2023 at 13:21
• @jd27 Ah I'm sorry for my misunderstanding. Now I got it. Thank you so much! Commented Aug 17, 2023 at 15:54

Using the suggestion of Severin Schraven i managed to solve the problem:

Since $$(X, \mathcal{A},\mu)$$ and $$(Y, \mathcal{B}, \nu)$$ are $$\sigma$$-finite there exist sequences $$(E_n)_{n \in \mathbb{N}}$$ and $$(F_n)_{n \in \mathbb{N}}$$ of increasing measurable sets each having finite measure with $$X = \bigcup_{n \in \mathbb{N}} E_n$$ and $$Y=\bigcup_{n \in \mathbb{N}} F_n$$.

Let $$f \in L^p(X \times Y)$$. Define a sequence $$(f_n)_{n \in \mathbb{N}}$$ of $$L^p(X \times Y)$$ functions by $$f_n = \chi_{E_n \times F_n} f$$. Now let $$(x,y) \in X \times Y$$. Then $$\lim_{n \to \infty } \chi_{E_n \times F_n} (x,y) = \lim_{n \to \infty } \chi_{E_n} (x) \chi_{F_n} (y) =1,$$ because there exist some $$m \in \mathbb{N}$$ so that $$x \in E_m$$ and $$y \in F_m$$ and then the same is true for all larger $$n$$ as well. Therefore $$f_n \to f$$ pointwise. Since $$|f_n| \leq |f |$$ the theorem of dominated convergence gives $$f_n \to f$$ in $$L^p$$.

Let $$\varepsilon>0$$ and $$k \in \mathbb{N}$$ so that $$\| f- f_k \|< \varepsilon$$. Now $$f_k$$ is zero outside of $$E_k \times F_k$$.

To proceed with the proof we need the following results: For a $$\sigma$$-Algebra $$\mathcal{A}$$ on a set $$X$$ and $$C \in \mathcal{A}$$ define the restricted $$\sigma$$-Algebra $$\mathcal{A}|_C$$ on $$C$$ by $$\mathcal{A}|_C = \{ A \cap C : A \in \mathcal{A} \}.$$ Now let $$\mu$$ be a measure on $$\mathcal{A}$$ then the map \begin{align} i: L^p(C, \mathcal{A}|_C, \mu ) & \longrightarrow L^p(X, \mathcal{A}, \mu ) \\ f & \longmapsto \bigg( x \mapsto \begin{cases} f(x), \quad \text{if} \ x \in C, \\ 0, \quad \text{else.} \end{cases} \bigg) \end{align} is a linear isometry, which can be easily verified on indicators and is then true for all (by density).

Now back to the proof: The measure spaces $$(E_k, \mathcal{A}|_{E_k}, \mu)$$ and $$(F_k, \mathcal{B}|_{F_k}, \nu)$$ are finite and furthermore $$\mathcal{A}|_{E_k} \otimes \mathcal{B}|_{F_k} =(\mathcal{A} \otimes \mathcal{B})|_{E_k \times F_k}$$ (see here).

By the finite case there exist a $$g \in L^p( E_k) \otimes L^p(F_k)$$ with $$\| g- f_k|_{E_k \times F_k} \|_{E_k \times F_k} < \varepsilon$$. Applying $$i$$ gives $$\|i(g) -f_k \|< \varepsilon$$. Furthermore $$i(g) \in L^p(X) \otimes L^p(Y)$$ and $$\|f-i(g) \| = \| f -f_k + f_k - i(g) \| \leq \| f-f_k\| + \|f_k- i(g) \| < 2 \varepsilon.$$

• Very nice proof. The only (irrelevant) detail that you could think about is that if you pick $f\in L^p(X\times Y)$ then you can (if we are pedantic) not talk about $f(x,y)$ unless you fix a representation. Commented Jul 18, 2023 at 16:34