I have a subset $B\subset\mathbb{R}^n\times\mathbb{R}^m$ that I want to show has measure zero. I know that the sections $B^x = \{y : (x,y)\in B\}$ all have measure zero. I do not know if $B$ is measurable. Is this enough to conclude that $B$ is a measurable set with measure zero?

EDIT : I would like to say that when randomly selecting $a\in\mathbb{R}^n\times\mathbb{R}^m$, I am almost always not in $B$. Since I know that $B^x = \{y : (x,y)\in B\}$ all have measure zero, it seems like a reasonable conclusion. I don't know enough probability or measure theory to put this in a rigorous way, so any suggestions would be great.

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    $\begingroup$ There is a counter-example in Rudin "Real and complex analysis", section 7.9 (page 143 in my edition). But this uses the continuum hypothesis, and the answers below seem to work in ZFC. $\endgroup$ Feb 24 '12 at 14:31

Since the Sierpinski article is in French an uses slightly old-fashioned notation, let me sketch a proof of the result.

Theorem. There is a function $f:\mathbb R\to\mathbb R$ whose graph is not a measurable subset of $\mathbb R^2$.

Proof. We first show that a set $A\subseteq\mathbb R$ of size $<2^{\aleph_0}$ cannot have a complement of measure zero (in fact, if such a set is measurable, then it is of measure zero, but we don't need this). To see this, we enlarge $A$ so that it becomes a subgroup of $(\mathbb R,+)$ that is still of size $<2^{\aleph_0}$.
Now choose $x\in\mathbb R\setminus A$. The coset $A+x$ is disjoint from $A$. Hence $\mathbb R$ is the union of $\mathbb R\setminus A$ and $\mathbb R\setminus(A+x)$.
It follows that not both of these sets can be of measure zero.
However, if $A$ is measurable, they have the same measure. So the complement of $A$ is not of measure zero.

Now let $(B_\alpha)_{\alpha<2^{\aleph_0}}$ be an enumeration of all Borel subsets of $\mathbb R^2$ of measure zero.
For each $\alpha<2^{\aleph_0}$ choose a pair $(x_\alpha,y_\alpha)\in\mathbb R^2$ such that $x_\alpha\not\in\{x_\beta:\beta<\alpha\}$ and $(x_\alpha,y_\alpha)\not\in B_\alpha$. This is possible since for each $\alpha$ the set $$((\mathbb R\setminus \{x_\beta:\beta<\alpha\})\times\mathbb R)\setminus B_\alpha$$ is not of measure zero by the remark at the beginning of the proof.

Now $\{(x_\alpha,y_\alpha):\alpha<2^{\aleph_0}\}$ is a partial function from $\mathbb R$ to $\mathbb R$ that is not contained in any measure zero Borel subset of the plane. It follows that this partial function is not of measure zero.
Extend the function to any total function $f:\mathbb R\to\mathbb R$.
The extended function still is not of measure zero. But since the sections are singletons, the graph of this function cannot be measurable by Fubini's theorem. \qed


Sierpinski gave an example of a nonmeasurable subset of $\mathbb{R}^2$ such that all sections are singletons and hence null sets. So the answer is in general no.


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