Linear combinations of indicator functions of measurable rectangles with sides of finite measure. The following is a question on measure theory whose answer would help me greatly in my research.

Let $ (X,\mathcal{F},\mu) $ and $ (Y,\mathcal{G},\nu) $ be $ \sigma $-finite measure spaces. Let $ E $ be a $ (\mathcal{F} \otimes \mathcal{G}) $-measurable subset of $ X \times Y $ such that $ (\mu \otimes \nu)(E) < \infty $.
Is it true that the indicator function $ \mathbf{1}_{E} $ is the pointwise limit of a sequence $ (s_{n})_{n \in \mathbb{N}} $ of functions on $ X \times Y $, where for each $ n \in \mathbb{N} $, we require that $ s_{n} $ be a finite linear combination of functions of the form $ \mathbf{1}_{A \times B} $, where $ A \in \mathcal{F} $, $ B \in \mathcal{G} $ and $ \mu(A),\nu(B) < \infty $? In other words, for each $ n \in \mathbb{N} $, we require that $ s_{n} $ be a finite linear combination of indicator functions of measurable rectangles whose sides have finite measure.

Thank you very much for your help.
 A: I believe it is true.

Notation
Let $ \mathcal{R} $ denote the collection of measurable rectangles in $ X \times Y $ with sides of finite measure.
Note: The null set $ \varnothing = \varnothing \times \varnothing $ is considered to be a measurable rectangle.
Let $ \mathcal{R}_{\sigma} $ denote the collection of subsets of $ X \times Y $ that are a countable union of elements of $ \mathcal{R} $.

Step 1
Fix $ \epsilon > 0 $, and let $ E $ be an $ (\mathcal{F} \otimes \mathcal{G}) $-measurable subset of $ X \times Y $ with finite measure. Then there exists an $ A \in \mathcal{R}_{\sigma} $ such that
$$
E \subseteq A
\quad \text{and} \quad
(\mu \otimes \nu)(A \setminus E) < \frac{\epsilon}{2}.
$$
This is part of the construction of the Carathéodory measure from a pre-measure (or any general set function). The Lebesgue measure on $ \mathbb{R} $ is constructed in this manner.
As $ A \in \mathcal{R}_{\sigma} $, we can find a sequence $ (R_{n})_{n \in \mathbb{N}} $ in $ \mathcal{R} $ such that $ \displaystyle A = \bigcup_{n = 1}^{\infty} R_{n} $. Then
$$
E \subseteq \bigcup_{n = 1}^{\infty} R_{n}
\quad \text{and} \quad
  (\mu \otimes \nu) \! \left( \bigcup_{n = 1}^{\infty} R_{n} ~ \Bigg\backslash ~ E \right)
< \frac{\epsilon}{2}.
$$
Note: We did not use the outer measure $ (\mu \otimes \nu)^{*} $ as we have assumed that $ E $ is measurable.

Step 2
Each $ A \in \mathcal{R}_{\sigma} $ is a countable union of disjoint measurable rectangles. Therefore, by Step 1, there is a sequence $ (D_{i})_{i \in \mathbb{N}} $ of disjoint measurable rectangles such that $ \displaystyle E \subseteq \bigcup_{i = 1}^{\infty} D_{i} $ and
$$
  (\mu \otimes \nu) \! \left( \bigcup_{i = 1}^{\infty} D_{i} ~ \Bigg\backslash ~ E \right)
< \frac{\epsilon}{2},
\quad \text{or equivalently}, \quad
  \left\| \chi_{\bigcup_{i = 1}^{\infty} D_{i}} - \chi_{E} \right\|_{L^{1}}
< \frac{\epsilon}{2}. \quad \quad (1)
$$

Step 3
Now, there exists an $ N \in \mathbb{N} $ such that
$$
  (\mu \otimes \nu) \!
  \left(
  \bigcup_{i = 1}^{\infty} D_{i} ~ \Bigg\backslash ~ \bigcup_{i = 1}^{N} D_{i}
  \right)
< \frac{\epsilon}{2},
\quad \text{or equivalently}, \quad
  \left\|
  \chi_{\bigcup_{i = 1}^{\infty} D_{i}} - \chi_{\bigcup_{i = 1}^{N} D_{i}}
  \right\|_{L^{1}}
< \frac{\epsilon}{2}. \quad (2)
$$
Combining $ (1) $ and $ (2) $, we have
$$
  \left\| \chi_{E} - \sum_{i = 1}^{N} \chi_{D_{i}} \right\|_{L^{1}}
= \left\| \chi_{E} - \chi_{\bigcup_{i = 1}^{N} D_{i}} \right\|_{L^{1}}
< \epsilon.
$$
As $ \epsilon $ is arbitrary, we can approximate $ \chi_{E} $ in the $ L^{1} $-sense by a sequence $ (s_{n})_{n \in \mathbb{N}} $, where each $ s_{n} $ is a finite sum of indicator functions of disjoint measurable rectangles whose sides have finite measure.
As there is a subsequence $ (s_{n_{k}})_{k \in \mathbb{N}} $ such that $ \displaystyle \lim_{k \to \infty} s_{n_{k}} = \chi_{E} $ pointwise almost-everywhere, we are done.
Note: Every $ L^{1} $-convergent sequence of functions contains a subsequence that converges to the very same limit pointwise almost-everywhere.
