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

Question

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.

Answer 1

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 $$\|f-i(g) \| = \| f -f_k + f_k - i(g) \| \leq \| f-f_k\| + \|f_k- i(g) \| < 2 \varepsilon.$$