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I already know that for Borel-$\sigma$-algebras it holds that $\mathfrak B^{p+q}=\mathfrak B^p \otimes\mathfrak B^q$. Now I want to show that this is not the case for Lebesgue-$\sigma$-algebras $\mathfrak L$.

So first of all, given a zero Lebesgue-null-set $N$ in $\mathbb R^p$, I have proven that $N\times B$ is a Lebesgue-null-set in $\mathbb R^{p+q}$ for arbitrary $B\subseteq \mathbb R^q$. Now, how can I show that $$\mathfrak B^p \otimes\mathfrak B^q\subsetneq \mathfrak L^p \otimes\mathfrak L^q \subsetneq \mathfrak L^{p+q} $$

My work for the first part of the proof: So if we chose $B$ where $\mu(B)<\infty$, it is obvious that we obtain $$\lambda(N\times B)=\nu(N)\mu(B)=0\times M=0$$ where $M\in\mathbb R$ is an upper bound of $\mu(B)$.

Thus I considered the case $\mu(B)≥\infty$. Since $\mu$ is $\sigma$ -additive, we can find a sequence $B_n\subseteq B$ such that $\bigcup_{n=1}^\infty B_n = B$ and $\mu(B_n)<\infty$ for all $n$. It follows that $$\lambda(N\times B)= \lambda(N\times \bigcup_{n=1}^\infty B_n )=\nu(N)\mu(\bigcup_{n=1}^\infty B_n) = \nu(N)\lim_{n\to\infty}\mu( B_n)=0\times S=0$$ where $S$ is the upper bound of $B_n$.

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I believe your proof solves completely the first part of the question. So I will only adress the second question:

$$\mathfrak L^p \otimes\mathfrak L^q \subsetneq \mathfrak L^{p+q}$$

The counterexample can be given for $p=q=1$. Then it can be extended for any indices.

Let $V$ be the Vitali set in $\mathbb{R}.$ We get that $N := V\times \{0\} \subset \mathbb{R} \times \{0\} \subset \mathbb{R}^2$ is contained in $\mathfrak L^{p+q},$ since it´s contained in a nullset.

But it´s not contained in the product Lebesgue sigma-algebra. This is because sections of sets in a product sigma-algebra must me measurable. But the section in $0$ of our set is not Lebesgue measurable, since $N_0 = V$ is the Vitali set.

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  • $\begingroup$ Thank you. What do you mean by "the section in $0$ of our set is not Lebesgue measurable"? What is the section in $0$? $\endgroup$
    – Tesla
    Commented May 26, 2016 at 18:58
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    $\begingroup$ If we have a set $E \subset X \times Y,$ for some sets $X$ and $Y$, define the section of $E$ in $x \in X$ as $E_x := \{ y \in Y | (x,y) \in E\}$. And similarly we can define the section of $E$ in $y \in Y.$ What you should know (or check up) is that a measurable set in a product sigma-algebra has measurable sections. $\endgroup$
    – Kore-N
    Commented May 26, 2016 at 19:16
  • $\begingroup$ Ok I was wondering, when I looked at my solution, how do we know that $\mathfrak L^p \otimes \mathfrak L^q \subset \mathfrak L^{p+q}$? $\endgroup$
    – Tesla
    Commented May 31, 2016 at 14:25
  • $\begingroup$ Take $A \times B $ in your product sigma-algebra $\mathfrak L^p \otimes \mathfrak L^q$. Then argue somehow that it actually has to lie in $\mathfrak L^{p+q}$. Now argue by monotone class argument. This is how I would intuitively do it. $\endgroup$
    – Kore-N
    Commented May 31, 2016 at 15:55
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    $\begingroup$ I'm not home now :) But in my mind it works like this. A lebesgue-mb set differs from a Borel set by a nullset. So write $A= A' \cup N, B = B' \cup M.$ Then $A \times B$ is union of sets that are either borel or have a nullset on at least one of the two sides. All these terms lie in the completition of the borel sigma-algebra of the product. $\endgroup$
    – Kore-N
    Commented May 31, 2016 at 16:22

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