Let $F$ be a local field of characteristic 0, and $f:F\rightarrow \mathbb{C}$ be an integrable function. Is the following formulation valid?
$ \int_{F^\times}f(x^2) d^\times x=\int_{F^{\times 2}}f(x) d^\times x $
where $d^\times x$ is a chosen multiplicative measure on $F^\times$.
It seems if $F=\mathbb{R}$, this is true. But is it true for any local field?
Since the subgroup of squares is open, the restriction of Haar measure from the full multiplicative group to the squares is a Haar measure on the subgroup of squares, and Haar measure is unique up to scalars.
This is false for non-archimedean local fields. On the other hand, as Paul Garrett explained, the two sides are equal up to an absolute constant.
Let $F=\mathbb{Q}_2$, and let $f$ be the characteristic function of $1+8\mathbb{Z}_2$. Then the left hand side is the measure of $1+2\mathbb{Z}_2$, while the right hand side is the measure of $1+8\mathbb{Z}_2$, hence they differ by a factor of $4$.
Let $F=\mathbb{Q}_p$ for $p>2$, and let $f$ be the characteristic function of $1+p\mathbb{Z}_p$. Then the left hand side is the measure of $\cup(\pm 1+p\mathbb{Z}_p)$, while the right hand side is the measure of $1+p\mathbb{Z}_p$, hence they differ by a factor of $2$.
I think for an appropriate choice of the measure on $F^{\times 2}$, I still believe this is true, which is essentially what Paul Garret mentioned. Indeed $F^{\times 2}\cong F^{\times}/ \{\pm1\}$, and the function $x↦f(x^2)$ can be viewed as a function on this quotient. Then \begin{equation} \int_{F^\times}f(x^2)d^\times x=\int_{F^{\times}/{\pm 1}}\sum_{\{\pm 1\}}f(y^2)d^\times y \end{equation}
where $d^\times y$ is the quotient measure. Then by identifying $F^{\times}/{±1}$ with $F^{\times 2}$, one can see that the left hand side is 2 times $\int_{F^{\times 2}}f(y)d^\times y$.
