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Is it provable in ZFC that there is a subset of the plane all of whose vertical cross sections have Lebesgue measure zero and all of whose horizontal cross sections are complements of sets of Lebesgue measure zero?

There are such sets in models in which every set of reals of cardinality less than the continuum is Lebesgue measurable.

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No. Suppose $m:\mathcal{P}([0, 1]) \to [0, 1]$ is a total extension of Lebesgue measure. Let $A \subseteq [0, 1]^2$ be such that every vertical section is Lebesgue null.

Claim: $A$ is $m \otimes \mu$-null.

Proof: For each rational $e>0$, let $U_{e, x} \subseteq [0, 1]$ be an open set of measure less than $e$ that contain the vertical section $A_x$. Let $G = \{(x, y) : (\forall e > 0)(y \in U_{e, x})\}$. For each rational interval $J$, let $X_{e, J} = \{x : J \subseteq U_{e, x}\}$. Then $G$ belongs to the sigma algebra generated by rectangles of the form $X_{e, J} \times J$ so $A$ is $m \otimes \mu$-null.

It follows that $\mu$-almost every horizontal section of $[0, 1]^2 \setminus A$ is not Lebesgue null. Note that this argument only requires the following: Every countably generated sigma algebra extending the Borel algebra admits a measure extending Lebesgue measure. A theorem of Carlson says that this holds in the random real model.

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  • $\begingroup$ I don't quite follow the last sentences. Suppose that, as you say, every countably generated $\sigma$-algebra extending the Borel $\sigma$-algebra admits an extension of Lebesgue measure. How do you get a total extension of Lebesgue measure? $\mathcal{P}([0,1])$ isn't countably generated. $\endgroup$ – Nate Eldredge May 29 '16 at 18:22
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    $\begingroup$ Also, what is $\mu$ here? Is it just Lebesgue measure? $\endgroup$ – Nate Eldredge May 29 '16 at 18:27
  • $\begingroup$ I added more details. The point is that for any $A$, there is a countable family of sets of reals such that whenever $m$ is an extension of Leb. measure that measure these sets, $A$ is $m \otimes \mu$-null ($\mu$ is Leb. measure). $\endgroup$ – Ashutosh May 29 '16 at 18:29
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    $\begingroup$ Thanks, Ashutosh. I thought that my question might be connected to measurable cardinals, but I did not see the neat argument you gave. $\endgroup$ – Bill Johnson May 29 '16 at 19:39
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In 1980 H. Friedman proved (in A consistent Fubini-Tonelli Theorem) that after adding $\mathfrak{c}^+$-many random reals the equality $\int f dx dy=\int f dy dx$ holds for any non-negative function for which the iterated integrals exist. In particular, a subset of the plane with the required properties does not exist in such model.

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  • $\begingroup$ Thanks, Ramiro. This explains why no real analysis textbook uses two dimensional Lebesgue measure to show that measurability is needed in Fubini/Tonelli to prove equality of the iterated integrals. $\endgroup$ – Bill Johnson May 30 '16 at 15:36
  • $\begingroup$ @BillJohnson: Could you provide some more details about your interesting comment? $\endgroup$ – Boaz Tsaban May 30 '16 at 15:40
  • $\begingroup$ Boaz, in real analysis texts there is usually a problem to show that measurability of $f(x,y)$ on a product space is necessary for proving that the iterated integrals $\int \int f(x,y)\, dx \, dy$ and $ \int \int f(x,y)\, dxy \, dx$ are the same when both exist and, say, $f \ge 0$. The examples always involve some "strange" measure space. Why not an example where both measures are Lebesgue measure on the line? You can do that under e.g. CH or MA--I alluded to that in my question. Friedman's result shows that you cannot do it just in ZFC. $\endgroup$ – Bill Johnson May 30 '16 at 18:38
  • $\begingroup$ @BillJohnson: Thanks for the clarification. Yes, that's very interesting and a bit surprising. $\endgroup$ – Boaz Tsaban Jun 14 '16 at 12:17

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