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Let $A,B \subset \mathbb{R}$ two bounded and Lebesgue measurable sets. I have to show that $A\times B \subset \mathbb{R}^2$ is measurable and \begin{align*} \lambda(A \times B) = \lambda (A) \cdot \lambda (B) \end{align*} where $\lambda$ is the Lebesgue measure.

I tried to show the equality of the measure using the Fubini's Theorem with the indicators function but I'm not sure is the right way and I have no idea how to show that $A \times B$ is measurable.

Any suggestions? Thanks in advance!!

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  • $\begingroup$ math.stackexchange.com/questions/2567276/… $\endgroup$ – user284331 Apr 20 '18 at 8:22
  • $\begingroup$ @user284331 I have only two measurable and bounded set in $\mathbb{R}$. Is there a way to say that they're also Borel sets? $\endgroup$ – userr777 Apr 20 '18 at 8:32
  • $\begingroup$ It should not be a problem to approximate the measure of a Lebesgue measurable set by a Borel measurable set. $\endgroup$ – user284331 Apr 20 '18 at 8:35
  • $\begingroup$ Of course not all Lebesgue measurable sets are Borel measurable, so do not assume them to be Borel one. $\endgroup$ – user284331 Apr 20 '18 at 8:35
  • $\begingroup$ @user284331 you mean writing e.g. $A = H \setminus N$ where $ H$ is a $G_{\delta}$-set (so Borel) and $N$ negligible set? $\endgroup$ – userr777 Apr 20 '18 at 8:42
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$A$ bounded is measurable if and only if for every $\epsilon >0$ there exists $U$ a finite union of intervals such that $m^{\star}(A\Delta U)<\epsilon $. Do this for $B$ too and get $$m^{\star}((A\times B) \Delta (U\times V))< 2M\epsilon $$ Multiplicativity should be easy now, since it works for $U$ and $V$.

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  • $\begingroup$ Not sure to have understood the notation; what do you mean with $ A \Delta U$ ? $\endgroup$ – userr777 Apr 21 '18 at 6:34
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    $\begingroup$ @userr777: en.wikipedia.org/wiki/Symmetric_difference $\endgroup$ – Orest Bucicovschi Apr 21 '18 at 6:52
  • $\begingroup$ I'm some trouble to get the inequality in your answer... you just used only the properties of the Cartesian product? $\endgroup$ – userr777 Apr 21 '18 at 7:21
  • $\begingroup$ @userr777:I also used that $A$, $B$ are each contained in a segment lf length $M$. $\endgroup$ – Orest Bucicovschi Apr 21 '18 at 7:37
  • $\begingroup$ To the proposer: The first sentence of this Answer is valid $only$ for a bounded set $A\subset \Bbb R.$ But $A\times B=\cup_{n\in \Bbb N}( (A\times B)\cap [-n,n]^2)$ so the main result follows from the result for bounded $A,B$. $\endgroup$ – DanielWainfleet Apr 21 '18 at 16:54

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