Let $\mu$ be a Haar measure on a locally compact Hausdorff topological group $G$, and let $H$ be a closed subgroup of $G$. If we restrict $\mu$ to the Borel sets of $G$ which are contained in $H$ (this restriction will include the Borel sets of $H$), it is easy to see that $\mu$ is a Borel measure on $H$. Let's call this restriction $\lambda$.

Sometimes $\lambda$ is a Haar measure on $H$, sometimes it isn't. If $G$ and $H$ are finite groups, then we can take $\mu$ and its restriction to be the counting measure. But if, for example, $G = \mathbb{R}$ and $H = \mathbb{Z}$, then the restriction of the Lebesgue measure to the Borel sets of $\mathbb{Z}$ (actually, the power set of $\mathbb{Z}$ since $H$ is discrete) is the zero measure.

I believe that if $H$ is a compact subgroup of $G$, then the only way that $\lambda$ can fail to be a Haar measure on $H$ is if it is the zero measure (that is, if $\mu(H) = 0$). Let me go through the properties of the Haar measure one by one, with the assumption that $H$ is a closed subgroup of $G$.

Translation invariance.

Obvious since $\mu$ is translation invariant.

Finite on compact sets.

Also clear, compactness transcends the subspace topology.

Inner-regular on open sets.

Let $V \cap H$ be open in $H$, for $V$ open in $G$. Actually, Haar measures are inner-regular on $\sigma$-finite Borel sets, so we just need $V \cap H$ to be $\sigma$-finite. This is the case when $H$ is compact.

Outer-regular on Borel sets.

If $E$ is a Borel subset of $H$, then $E$ is also a Borel subset of $G$. Let $V$ run through the open sets of $G$ which contain $E$, or equivalently for which $E \subseteq V \cap H$. We have $$\mu E = \inf_V \mu(V) \geq \inf_V \mu(V \cap H) \geq \mu(E)$$

At a glance, can anyone tell if I've made any mistake here?


Steinhaus-like theorems (see, here, here, here, here) imply that if $H$ is a closed subgroup of a locally compact topological group and Haar measure of the group $H$ is positive then $H$ is open.

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    $\begingroup$ So inner-regularity follows from there, because open subsets of $H$ are also open in $G$. Thank you. $\endgroup$ – D_S Jun 24 '15 at 13:49

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