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Is the following slightly generalization of Corollary 20.17 in Hewitt and Ross Book (page 296) correct?

Let $A$ be a subset of a profinite group $G$ ( compact, Hausdorff, totally disconnected topological group) such that $A=A^{-1}$ and $\lambda(A)>0$, where $\lambda$ is the normalized Haar measure of $G$. Suppose that $f$ is the function defined on $G\times \cdots \times G$ ($k$ times) into $\mathbb{R}^{\geq 0}$ by $$(x_1,\dots,x_k)\mapsto \lambda(A\cap x_1 A \cap \cdots \cap x_k A).$$ Then $f$ is continuous.

Note that $f(1,\dots,1)=\lambda(A)$ ( $1$ is the identity element of $G$), so $f$ is somewhere positive.

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    $\begingroup$ I bet it is: if $x_{j,n}$ converges to $x_j$, then $\mathbb{1}_{x_{j,n}A}$ converges a.e. to $\mathbb{1}_{x_jA}$, and thus $\mathbb{1}_{A \cap x_{1,n}A \cap \ldots \cap x_{k,n}A}$ converges a.e. to $\mathbb{1}_{A \cap x_{1}A \cap \ldots \cap x_{k}A}$. The desired result follows now by the dominated convergence theorem. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Nov 15, 2019 at 9:03
  • $\begingroup$ @MateuszKwaśnicki: I think your first claim will be true if $A$ is open, am I right? $\endgroup$
    – MSMalekan
    Commented Nov 15, 2019 at 11:06
  • $\begingroup$ @MeisamSoleimaniMalekan: I meant general Borel $A$, but I was indeed too sketchy when I wrote "converges a.e. to $\mathbb{1}_{x_jA}$": what I should have written is "converges in measure to $\mathbb{1}_{x_jA}$, and hence every subsequence has a sub-subsequence convergent a.e.". $\endgroup$
    – Mateusz Kwaśnicki
    Commented Nov 15, 2019 at 11:35

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The answer is yes, and the assumption of profiniteness is not needed (see comments). This can be proved by approximating in $L^2$ the indicator function $1_A$ by a continuous function $f$.

Using the fact that such $f$ must be uniformly continuous, any small perturbation $f_g$ of $f$ is within a small $L^\infty$ error of $f$. Therefore the map $$\Phi_f:(x_1,\dots,x_k)\mapsto\int_G f(g)f(x_1^{-1}g)\cdots f(x_k^{-1}g)\,d\lambda(g)$$ is continuous.

As $f\to1_A$ in $L^2$, Cauchy-Schwarz's inequality (together with the fact that we can keep every function in this argument bounded by $2$) implies that $\Phi_f\to\Phi_{1_A}$ uniformly, which shows that $\Phi_{1_A}$ is also continuous. Finally, note that $$\Phi_{1_A}(x_1,\dots,x_k)=\lambda(A\cap x_1A\cap\cdots\cap x_kA).$$

The more general principle at play here is that for any $p<\infty$ the map $g\mapsto f_g$ from $G$ to $L^p(G)$ is continuous, where $f_g$ denotes the function $f_g(x)=f(gx)$.

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    $\begingroup$ I guess "assumption of profiniteness is not needed" has to be interpreted as: being, instead, locally compact ($\sigma$-compact?) is enough. $\endgroup$
    – YCor
    Commented Nov 18, 2019 at 11:08
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    $\begingroup$ @YCor I was thinking of compact Hausdorff (I should have said totally disconnected is not needed) but now that you mention it I think locally compact should be enough. All that's needed is existence of Haar measure and denseness in $L^2$ of compactly supported continuous functions. $\endgroup$
    – Joel Moreira
    Commented Nov 18, 2019 at 15:16
  • $\begingroup$ But you deduce uniform continuity of $f$ from continuity of $f$, so surely need compactness? (EDIT: Oops, sorry, I just saw 'compactly supported'.) $\endgroup$
    – LSpice
    Commented Nov 18, 2019 at 16:24

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