Restriction of Haar Measure Induces a Restriction of Integral Formula

Question

I just need some reality check. Suppose we have a locally compact abelian group $G$ with an open subgroup $H$. We equip $G$ with a Haar measure $\mu_G$, while we equip $H$ with the Borel measure $\mu_H$ induced by restricting $\mu_G$ to Borel sets of $H$. It turns out that by open-ness of $H$, we are guaranteed that $\mu_H$ is also a Haar measure. Is it true that for any $f\in L^1(G,\mu_G),$ $$\int_G \chi_{H}(x)f(x)d\mu_G(x) = \int_{H}f_{| H}(x) d\mu_H(x)$$ given that $\chi_H$ is the indicator function on $H$, and $f_{|H}$ is the restriction of $f$ on $H$? I have a feeling that the formula above is true regardless of the fact that $H$ is open. In the case that it is not, it's just that we are not guaranteed that $\mu_H$ has the necessary properties to be a Haar measure (I think outer regularity won't be automatic).