Suitable non-countable union of sets with measure zero is still a set of measure zero?

Question

Let $R=[0,a]\times [0,b]$ be a rectangle in $\Bbb{R}^2$. Associated to each $t\in [0,a]$ we have a set $E_t\subset [0,b]\subset \Bbb{R}$ which has measure zero (it is worth to remark that here we are considering the Borel measure of the real line, or "length", if we want). Set $\mathscr{E}_t=\{t\}\times E_t\subset R\subset \Bbb{R}^2$. So, my question is: still has the set

$$\bigcup\limits_{t\in [0,a]}\mathscr{E}_t\subset R\subset \Bbb{R}^2 $$

measure zero? (considering now the measure of the plane, or "area").

If this is true, is there some theorem or result that garantees it easily? If this is false, which is the counterexample?

THANK YOU, GUYS!

Answer 1

Let us denote your set by $E$, Lebesgue measure on $\mathbb{R}$ by $\mu_1$, and Lebesgue measure on $\mathbb{R}^2$ by $\mu_2$. If $E$ is measurable, then by Fubini's theorem applied to the characteristic function of $E$, $$\mu_2(E)=\int_{[0,a]}\mu_1(E_t)dt=0.$$ More generally, this shows that the inner measure of $E$ is always $0$, by applying this argument to any measurable subset of $E$.

However, $E$ need not be measurable. In fact, it is possible to choose sets $E_t$ such that each one has at most one point, but the outer measure of $E$ is $ab$, the measure of the entire rectangle. You can prove this by transfinite induction: there are only $\mathfrak{c}$ different Borel subsets of $[0,a]\times[0,b]$ of positive measure and each of them intersect $\{t\}\times[0,b]$ for $\mathfrak{c}$ different values of $t$, and so by an induction of length $\mathfrak{c}$ you can choose at most one point from each $\{t\}\times[0,b]$ and get a set that intersects every such Borel set.

Answer 2

If $E=\bigcup_t \mathcal E_t$ is measurable, then the answer is yes, directly by Fubini's theorem. In general, (assuming axiom of choice) it is not true, by a construction similar to that of the Bernstein set.

Let $(F_\alpha)_{\alpha<\mathfrak c}$ be an enumeration of all closed subsets of $R$ of positive measure. By easy transfinite induction you can construct a sequence $(t_\alpha,s_\alpha)_{\alpha<\mathfrak c}$ such that $(t_\alpha,s_\alpha)\in F_\alpha$, and such that for $\alpha\neq \beta$ we have $t_\alpha\neq t_\beta$ (this can be done because the projection of a closed set of positive measure onto one axis will always have size continuum -- for example by Fubini).

Then put $E_t=\{s_\alpha\}$ if $t=t_\alpha$ and $E_t=\emptyset$ if for no $\alpha$ we have $t=t_\alpha$. Clearly, each $E_t$ is null, but $\bigcup_t \mathcal{E}_t$ has full outer measure: indeed, by construction it intersects every closed set of positive measure, which -- as you can easily check -- is equivalent to having full outer measure.