Let $E\subseteq\Bbb R$ be measurable with positive Lebesgue measure. Let $E_1,E_2$ be the projections of $E$ on the $x$ – axis and $y$ – axis respectively. Can we say that $E_1,E_2\subseteq\Bbb R$ are measurable with positive Lebesgue measures?
Projection of a measurable set of positive measure.
lebesgue-measuremeasurable-functionsmeasure-theory
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I'll expand copper.hat's point into an answer. To prove that the product of measurable sets in $\mathbb{R}^d$ is measurable, it suffices to show that the product of measurable set of finite measure in $\mathbb{R}^d$ is measurable (this generalizes to arbitrary $\sigma$-finite measure spaces).
Proof: Let $E_{1},E_{2}$ be given as above. Define $E_{1,N} = E_{1} \cap B(0,N)$, the intersection of $E_{1}$ with the ball of radius $N$. This is still measurable, as it is the intersection of two measurable sets, and it is has finite measure by monotonicity of measure, as $B(0,N)$ has finite measure. Similarly define $E_{2,N}$. By hypothesis, we have proved that $E_{1,N} \times E_{2,N}$ is measurable for any choice of $N$. But now we note that
$$E_{1} \times E_{2} = \bigcup_{N \in \mathbb{N}}(E_{1,N} \times E_{2,N})$$
So $E_1 \times E_2$ is the countable union of measurable sets, and hence measurable.
You showed that $l^*(E) = \inf l^*(\mathcal{O})$ (the infimum runs over all open sets $\mathcal{O}$ containing $E$).
Suppose $E$ is of finite exterior measure -then for any $n\in \mathbb{N}$ we have an open set $\mathcal{O_n}$ such that $$l^*(\mathcal{O_n}) \leq l^*(E)+n^{-1}<\infty$$ And since $l^*(\mathcal{O_n}) = l^*(E)+l^*(\mathcal{O_n} \setminus E)$ we have $l^*(\mathcal{O_n} \setminus E) \leq n^{-1}$ (The equality is infact a restatement of the measurability of $E$, see Equivalent Definition of Measurable set)
Taking $\mathcal{O} = \bigcap_{n=1}^\infty \mathcal{O_n}$ we have come up with a $G_\delta$ set (countable intersection of open sets) which is obviously a borel set, with the property: $$l^*(\mathcal{O}\setminus E) \leq l^* (\mathcal{O_n}\setminus E) \leq n^{-1}$$ for all $n\in\mathbb{N}$ which is the first corollary.
If $E$ is of infinte exterior measure then we denote $E_n = E \cap B(0,n)$ ($E$'s intersection with the ball of radius $n$ centered at the origin). Each $E_n$ is bounded and so has finite exterior measure (Since it obviously is encompassed within $B(0,n)$) And we may extract $G_n$ sets such that $$l^*(G_n \setminus E_n)=0$$ Putting $G=\bigcup_{n=1}^\infty G_n$ we have $$l^*(G\setminus E) = l^*\left(\bigcup_{n=1}^\infty G_n \setminus \bigcup_{n=1}^\infty E_n \right)\leq \sum_{n=1}^\infty l^*(G_n \setminus E_n ) =0$$
This shows the first corollary.
For the second: For every measurable $E^C$ we obviously have a borel set $G\supset E^C$ such that $l^*(G\setminus E^C) =0$ (Take the set from the first claim) that shows that $G\setminus E^C$ is a lebesgue measurable set as it is a null set. Then obviously $E^C= G \setminus (G\setminus E^C)= G \cap (G\cap E)^C$. Taking completements on both sides we have:$$ E = G^C \cup (G\setminus E^C)$$
$G^C$ is again a borel set (As a completement of borel set).
As a final comment on this excercise: One notices we used 2 main properties of the Lebesgue measure:
- It is outer regular meaning that for every measurable subset $E$ , for every $\varepsilon>0$ we have an open set $\mathcal{O}\supset E$ with the property $l^*(\mathcal{O} \setminus E) \leq \varepsilon$ which we had to prove.
- It is defined on borel sets, bounded sets have finite exterior-measure.
It turns out that the second implies the first and the first property is what we actually needed. In this proof I made a slight detour (as i wasn't sure which Lebesgue measurability criterion you are using).
Best Answer
Projections need not be measurable, but if they are measurable then their measures are necessarily positive: if $A$ and $B$ are the projections of $E$, then $E \subset A\times B$ and the measure of $A\times B$ would be $0$, if one of the projections has measure $0$.
Counterexample: let $A$ be any non-measurable set in $\mathbb R$ and $E=A\times \{0\}$. Then $E$ is Lebesgue measurable but its first projection is not. If you want to get an example where $E$ has positive measure but one of its projections is not measurable, just take the union with a disjoint rectangle; for example, we can take $A$ to be a non-measurable subset of $[0,1]$ and consider $(A\times \{0\}) \cup ([2,3]\times [2,3])$.