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Let $E$ be a set in $R^n$ with $m^*(E) >0$and let $\theta$ satisfy $0<\theta<1$. Show that there is a set $E_θ \subset E$ with $m^*(E_\theta) = \theta m^*(E)$ and that $E_\theta$ can be chosen to be measurable if E is measurable.

My try:

Define the function $f(x)= m^*(E \cap Q(x))$ where $Q(x)$ denotes the cube with edge $x$ centered at the origin(with vertices all of whose coordiantes $\pm \frac{x}{2}$ and $0<x< \infty$)

Observe that this function is continous and increasing.(why?)

Given that, observe que $\lim_{x \to 0}f(x) = 0$ and $\lim_{x \to \infty }f(x) = m^*(E)$.

That implies that there is a point $x_0$ such that $$f(x_0)=\theta\cdot m^*(E)$$ Then $E_\theta=E \cap Q(x_0)$.

My questions:

  • Is my proof correct?
  • How to prove that $f(x)$ is continous?
  • Concluding that $E_\theta$ can be chosen to be measurable if $E$ is measurable, it is the same as saying that for our construction of $E_\theta$ implies to be measurable because the intersection of two measurable sets is measurable?
Rodrigo Palacios
  • 323

1 Answers1

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For points 1,3) your proof is correct and your remark about measurability is correct.

For point 2), it is not too hard. For $h\gt0$:

$$\begin{align}m^\ast(Q(x+h)\cap E)&\le m^\ast((Q(x+h))\setminus Q(x))\cap E)+m^\ast(Q(x)\cap E)\\\implies m(Q(x+h)\setminus Q(x))&= m^\ast(Q(x+h)\setminus Q(x))\\&\ge m^\ast((Q(x+h)\setminus Q(x))\cap E)\\&\ge m^\ast(Q(x+h)\cap E)- m^\ast(Q(x)\cap E)\\&=f(x+h)-f(x)\\&\ge 0\end{align}$$And since $m(Q(x+h)\setminus Q(x))=(x+h)^n-x^n\to0,\,h\to0^+$ and since we have a symmetric inequality for $h\lt 0$ with the same conclusion, we must have that $\lim_{h\to0}|f(x+h)-f(x)|=0$, that is, $\lim_{h\to0}f(x+h)=f(x)$ and $f$ is a continuous function on $\Bbb R$.

You may be interested in a more general fact, the Sierpinski continuum of measure theorem, of atomless measure spaces. Since the real Lebesgue measure spaces are atomless (why?) Sierpinski's theorem gives your result immediately. That said, the standard proof that the Lebesgue measure is atomless uses a procedure similar to your proof.

See here and here for reference on this continuum of measure.

FShrike
  • 40,125