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Let $\mu$ be a finite Borel measure on $\mathbb R$, i.e. a finite measure on the Borel $\sigma$-algebra $S (\mathbb R)$, and let $B$ be a Borel subset of $\mathbb R$. Define the function $f$ on $\mathbb R$ by $f (x) =\mu (B + x)$.

(a) Show that $f$ is Borel measurable. (b) Show that $\int f (x) d\lambda (x) =\int f (x) dx =\mu(R)\lambda(B)$, where $\lambda$ denotes the Lebesgue measure

I am not able to prove that $f$ is Borel measurable. I tried to prove that $f^{-1}(a,\infty)$ is a Borel set but couldn't prove it.

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  • $\begingroup$ Can you do it in case $B$ is an open interval? $\endgroup$ – GEdgar Dec 16 '13 at 18:34
  • $\begingroup$ I tried that but I still couldn't do it. $\endgroup$ – Oliver Dec 16 '13 at 18:36
  • $\begingroup$ A still easier case is $B=(a,b)$ but also $\mu(\{x\}) = 0$ for all $x$. Can you prove $f$ is Borel measurable in that case? $\endgroup$ – GEdgar Dec 16 '13 at 18:38
  • $\begingroup$ I cant't do that either. $\endgroup$ – Oliver Dec 16 '13 at 19:40
  • $\begingroup$ I was able to get part a). Can some one help me with part b) $\endgroup$ – Oliver Dec 17 '13 at 0:06
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For the second part, notice that $f(x)=\int \chi_B(s-x)\mathrm d\mu(s)$. We then use Fubini's theorem.

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Use the product measure $\mu\times\lambda$ in $\mathbb R^2$ and integrate the clearly $\mu\times\lambda$-measurable function $F(x,y)=\chi_B(x-y)$. Fubini's Theorem answers both question. (See the version of Fubini's Theorem in Rudin's Real & Complex Analysis.)

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