To find a measurable subset with arbitrary measure

Question

Suppose $E$ is measurable subset of $\Bbb R$ s.t. $m(E)=1$. Does there exist $A$ that is measurable subset of $E$ and $m(A)=\frac 1 2$?

$A\subset E$ so $m(A) \le m(E)=1$. since $0 \le m(A)\le 1$, then $m(A)$ can be $\frac 1 2$

The question is: if this statement is true, how to prove it (is enough this reason?) If false what's a counterexample.

Answer 1

I wouldn't say this differs very much from the already posted solution, but may make a little bit more sense for you. Notice that you can't solve the problem without dealing with the concept of continuity (the continuity of Lebesgue measure for example).

So, using the continuity of Lebesgue measure we get that the function $f:\mathbb{R}\to[0,1]$ defined by $$f(x) = m\{t\in E : t\le x\}=m\left(E\cap(-\infty,x)\right)=m E_x,\quad x\in\mathbb{R}$$ is continuous. (Note that $E_x\subset E_y$ for $x\le y$ from which you can conclude that $f$ is increasing. However, it is not a neccesary part). From the definition of $f$ (and the continuity of measure) follows $$\lim_{x\to -\infty}f(x)=\lim_{x\to -\infty}m\left(E\cap(-\infty,x)\right)=m(\emptyset)=0,$$ $$\lim_{x\to \infty}f(x)=\lim_{x\to\infty}m\left(E\cap(-\infty,x)\right)=m(E)=1.$$

Now, from the Intermediate value theorem we can claim that $f(c)=1/2$ for some $c\in\mathbb{R}$, which will give us the set $A=\{t\in E : t\le c\}=E\cap(-\infty,c)\subset E$ of measure $1/2$: $$m(A)=m\{t\in E : t\le c\}=f(c)=\dfrac{1}{2}.$$

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Another way to show the continuity of $f$ is the following: as the set $E$ measurable then the characteristic function $\chi_E$ being bounded is also measurable, namely, it is summable, hence the function $$f(x) = \int_{-\infty}^x \chi_E(t)\,dt,\quad x\in\mathbb{R}$$ is (absolutely) continuous.

Answer 2

Hint: Consider intersections of $E$ with intervals, and use the Intermediate Value Theorem.

Answer 3

Elaboration of the hint assuming $m$ is Lebesgue measure: Define $f :\mathbb{R}_{\geq 0}\rightarrow \mathbb{R}$ by $f(x)=m(E_x)$ where $E_x=E\cap (-x,x)$. As $E_n \nearrow E$, $\lim_{x\rightarrow \infty}f(x)=m(E)=1 $. Also $f(0)=0$. So if we can show that $f$ is continuous by intermediate value theorem we are done. To show continuity, let $(x_n)$ be a sequence converging to $x$. Then note that $f(y)=\int \chi_E (t)\chi_{(-y,y)} (t) dm(t)$. Hence by dominated convergence theorem, $(x_n)$ converges to $x$ implies $f(x_n)$ converges to $f(x)$.

I think this proof works for any measure as long as we assume singleton sets have zero measure.

If $m$ is arbitrary it need not be true as consider $\delta$ on $2^\mathbb{R}$ so that $\delta (E)=0 $ if $0$ is not in $E$ and $\delta (E)=1$ if $0\in E$.