Can we show $\mu(gX) = \mu(X)$ if integral is translation invariant? Let $\mu$ be a Radon probability measure on the compact Hausdorff topological group $G$ such that
$$\int_G f(g) \mu(dg) = \int_Gf(hg) \mu(dg)$$
for all $h \in G$ and for all $f \in C(G)$.
Can I deduce that $\mu(hX) = \mu(X)$ for a Borel set $X$ and  $h \in G$?
Attempt: 
$$\mu(hX) = \int_G I_{hX}(g) \mu(dg) = \int_G I_{X}(h^{-1}g) \mu(dg)\stackrel{(?)}= \int_G I_{X}(g) \mu(dg) = \mu(X)$$
Now, I try to explain $(?)$. Maybe I can approximate with continuous functions or something like that?
 A: Observe that it is enough to prove this result for any compact set $X$.
You have $I_{hX}\in L^1(G)$.
Now $C_c(G)$ is dense in $L^1(G)$
then $\exists\{f_n\}$ in $C_c(G)$ such that $f_n\to f$ in $L^1$ norm.
$$\|f_n-I_{hX}\|_1\xrightarrow[]{n\to \infty} 0\Leftrightarrow\int_G |f_n(g) -I_{hX}(g)| \mu(dg)\to 0$$$$\implies \left|\int_G f_n(g)\mu(dg) -\int_GI_{hX}(g) \mu(dg)\right|\to 0$$$$\implies \lim_n \int_G f_n(g)\mu(dg)=\int_GI_{hX}(g) \mu(dg)=\mu(hX)$$
It is always possible to choose this sequence $\{f_n\}$ in such a way that for each  $n$, $0\le f_n\le 1$ and $f\equiv1$ on $X$.(refer here)
As $f_n\xrightarrow[]{L_1}I_{hX}$ there is a sub sequence of $\{f_{n_k}\}$ such that $f_{n_k}\to I_{hX}$ almost everywhere in $G$.
Then we have $f_{n_k}(hg)\to I_{hX}(hg)$ for almost every $g\in G$. As we have $|f_{n_k}|\le1$ Dominated convergence theorem applies here. 
So we get
 $$\int_{G}|f_{n_k}(hg)- I_{hX}(hg)|\mu(dg)\xrightarrow[]{k\to \infty}0$$
$$\implies\left|\int_G f_{n_k}(hg)\mu(dg) -\int_GI_{hX}(hg) \mu(dg)\right|\to 0 $$$$\implies \lim_n \int_G f_{n_k}(hg)\mu(dg)=\int_GI_{hX}(hg) \mu(dg)$$$$\implies\lim_n \int_G f_{n_k}(g)\mu(dg)=\int_GI_{X}(h^{-1}hg) \mu(dg)=\int_GI_{X}(g) \mu(dg)=\mu(X)$$
Hence we have $\mu(hX) = \mu(X)$.
Warning: This proof only works for $G$ Locally compact Hausdorff Group. 
