Let $A$ be a Lebesgue measurable subset of $\mathbb{R}$ and let $C_A=\{(x,y)\in\mathbb{R}^2;x\in A\}$. Such sets are called cylinder sets.

(i) Show that the collection of these sets forms a ring $R_C$.

(ii) Show that the set function $\mu(C)$ defined by $\mu_C(C_A)=\mu_L(A)$ is a measure on this ring.

(iii) Show that if $S$ is a proper subset of $\mathbb{R}$ and $\mu_L(A)\neq 0$, the set $A\times S=\{(x,y);x\in A,y\in S\}$ is not a measurable subset of $\mathbb{R}^2$ with respect to the measure $\mu_C$. (Hint: What is its outer measure, computed with respect to $\mu_C$?)

For (i) the collection of those sets form a ring because the collection of Lebesgue measurable sets on $\mathbb{R}$ form a ring.

(ii) follows trivially.

Now, I don't understand the hint in (iii). The outer measure should be $\mu_L(A)$, but why does that prove $A\times S$ is not a measurable subset?

  • $\begingroup$ And what about the complement? $\endgroup$ – Etienne Sep 15 '13 at 15:01
  • 1
    $\begingroup$ @Etienne Yeah I think I got it just before you posted. The complement also have measure $\mu_L(A)$, and so you need $\mu_L(A)+\mu_L(A)=\mu_L(A)$, not true! $\endgroup$ – PJ Miller Sep 15 '13 at 15:02

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