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Let $m$ be the Lebesgue measure on $[0,1)$, which is the fundamental domain of $\mathbb{R}/\mathbb{Z}$. Consider the bijection $\iota:\mathbb{R}/\mathbb{Z}\to[0,1)$ and define a measure $\mu$ on $\mathbb{R}/\mathbb{Z}$ with $\mu(E)=m(\iota(E))$ for any Borel set $E\subset\mathbb{R}/\mathbb{Z}$. I would like to show that $\mu$ is a Haar measure. But I'm stuck with showing the inner regularity of $\mu$ (yes, I would like to show directly that $\mu$ is a Radon measure).

For instance, let $E =[0.2, 0.8] \hbox{ mod } 1$. Then clearly by the inner regularity of Lebesgue measure, one has

$$m(\iota(E))=\sup\{m(K):K\subset \iota(E), K\textrm{ compact in } [0,1)\}.$$

However, how can I show that

$$m(\iota(E))=\sup\{m(\iota(K))\:K\subset E, K\textrm{ compact in } {\mathbb R}/{\mathbb Z}\}?$$

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  • $\begingroup$ More generally, you can show that if $f: X \to Y$ is continuous and $\mu$ is an inner regular borel measure on $X$, then $f_{*} \mu$ is also inner regular (where $f_{*} \mu$ is the pushforward measure). In your case, as D_S pointed out, you would consider the projection $f : [0,1) \to \mathbb{R}/\mathbb{Z}$ (rather than $\iota$). $\endgroup$ – Dominique R.F. Mar 9 '17 at 4:49
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All you really need is the inner regularity of Lebesgue measure.

The bijection $i: \mathbb{R}/\mathbb{Z} \rightarrow [0,1)$ is not continuous, but its inverse is. Hence the preimage of any compact set in $[0,1)$ is compact in $\mathbb{R}/\mathbb{Z}$. On the other hand, $K$ could be a compact set in $\mathbb{R}/\mathbb{Z}$, but $\iota(K)$ need not be compact.

If $E \subseteq \mathbb{R}/\mathbb{Z}$, then the (Lebesgue) measure $s := m(\iota(E))$ of $\iota (E)$ is the supremum of $m(C)$, as $C$ runs through all compact subsets of $\mathbb{R}$ which are contained in $\iota(E)$. Such compact sets are necessarily contained in $[0,1)$, so $ m(\iota(E))$ is the supremum of $m(C)$, as $C$ runs through all compact sets of $[0,1)$ which are contained in $\iota(E)$.

Let $S = \sup \{ m(\iota(K)) : K \subseteq E \textrm{ and is compact } \}$. If $C \subseteq \iota(E)$ is compact, then $K = \iota^{-1}(C)$ is compact in $\mathbb{R}/\mathbb{Z}$ and contained in $E$, with $m(C) = m(\iota(K))$, so certainly $S \geq s$ (you are taking a supremum over more things).

To show that $S$ does not strictly exceed $s$, it suffices to show that for any compact set $K$ contained in $E$, the measure of $\iota(K)$ can be approximated from below by compact sets in $[0,1)$ which are contained in $E$. But this is clear, because $\iota(K)$ can be approximated by compact sets in $[0,1)$ which are contained in $\iota(K)$ (hence in $E$).

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