definition of a vector-valued integral of elements of $L^\infty$ Let $\phi \in L^\infty(\mathbb{R}^n)$. I would like define an element $\tilde\phi$ of $L^\infty(\mathbb{R}^n)$ by the integral 
$$
\tilde\phi(x)=\int_{SO_n(\mathbb{R})} \phi(Ax) d\mu(A)
$$
where $\mu$ is the normalized Haar measure of the group $SO_n(\mathbb{R})$ of special orthogonal transformations and I would like write for any $f \in L^1(\mathbb{R}^n)$ something as
$$
f(\cdot)\int_{SO_n(\mathbb{R})} \phi(A\cdot) d\mu(A)
=\int_{SO_n(\mathbb{R})} f(\cdot)\phi(A\cdot) d\mu(A)
$$
How define the integral the $L^\infty(\mathbb{R}^n)$-valued integral $\int_{SO_n(\mathbb{R})} \phi(A\cdot) d\mu(A)$ ?
I try to use the Bochner integral but the function $A \mapsto\phi(A\cdot)$ does not seem measurable.
 A: For some functions $\phi$ the map $C_\phi:SO(n)\to L^\infty(\Bbb R^n)$, $A\mapsto \phi\circ A$ is continuous so certainly measurable. For the moment lets look at how this case would define an element $\int_{SO(n)}C_\phi(A) d\mu$ of $L^\infty(\Bbb R^n)$.
Since $\|C_\phi(A)\|=\|\phi\|$ and $SO(n)$ has finite volume $\int_{SO(n)}\|C_\phi(A)\|d\mu<\infty$. It follows for any $f\in L^1(\Bbb R^n)$ that $\langle f,C_\phi(A)\rangle$ is an integrable function $SO(n)\to\Bbb C$, so the linear map
$$\Psi:L^1(\Bbb R^n)\to\Bbb C,\ f\mapsto\int_{SO(n)}\langle f,C_\phi(A)\rangle d\mu\tag{1}$$
is well defined. It is also continuous since
$$|\Psi(f)|≤\int|\langle f,C_\phi(A)\rangle d\mu≤\|f\|\int_{SO(n)}\|C_\phi(A)\|d\mu = \|f\|\,\|\phi\|\cdot \mathrm{Vol}(SO(n))$$
so in this way we have defined an element of $(L^1(\Bbb R^n))^*=L^\infty(\Bbb R^n)$. This element is $\int_{SO(n)}C_\phi(A) d\mu$.
Now I don't know whether the map $C_\phi$ is measurable for all $\phi$ (it certainly isn't continuous for all $\phi$), but we can ask if the definition of $\Psi =\int_{SO(n)}C_\phi(A)d\mu$ via $\Psi(f):=\int_{SO(n)}\langle f,C_\phi(A)\rangle d\mu$ works anyway to define an element of $L^\infty(\Bbb R^n)$. In the case where $C_\phi$ is measurable this definition is just the normal definition, so this would be an extension of the usual integral.
First abuse notation to define $C_f(A)$ for $f\in L^1(\Bbb R^n)$ and now note that:
$$\langle f, C_\phi(A)\rangle = \int_{\Bbb R^n}d^nx\, f(x)\phi(Ax)=\int_{\Bbb R^n}d^nx\,f(A^{-1}x)\phi(x)=\langle C_f(A^{-1}),\phi\rangle$$
if we can show the function $A\mapsto C_f(A^{-1})$ is measurable it follows that the function $A\mapsto \langle C_f(A^{-1}),\phi\rangle$ is an integrable function $SO(n)\to\Bbb C$, since $|\langle C_f(A^{-1},\phi\rangle|≤\|f\|\,\|\phi\|$ and $SO(n)$ has bounded volume. It would follow that the prescription $(1)$ is well defined and we get our vector $\Psi = \int_{SO(n)}C_\phi(A)d\mu$.

The map $A\mapsto C_f(A^{-1})$ is actually continuous. To see this let $A_k\to\Bbb 1$ in $SO(n)$, we would like to see why
$$\|C_f(A_k^{-1})-C_f(\Bbb1)\|=\int_{\Bbb R^n}d^nx\,|f(A_k^{-1}x)-f(x)|$$
converges to zero. Here the trick is to use that
$$\lim_{R\to\infty}\int_{\Bbb R^n- B_R(0)}d^nx\,|f(x)| = 0$$
and that the $A_k$ are rotations to get that for a fixed $f$ and for any $\epsilon$ there is an $R$ so that
$$\int_{\Bbb R^n-B_R(0)}d^nx\,|f(A^{-1}_kx)-f(x)|<\epsilon/2$$
if we can show that there exists an $N$ so that if $k>N$ $\int_{B_R(0)}d^nx\,|f(A_k^{-1}x)-f(x)|<\epsilon/2$
then we've got the convergence. Let $\delta_k :=\sup_{x\in B_R(0)}\|A^{-1}_kx-x\|$ be the maximal amount that $A^{-1}_k$ translates a vector in the ball $B_R(0)$. You've got
$$\int_{B_R(0)}d^nx\ |f(A^{-1}_kx)-f(x)|≤ \int_{B_{\delta_k}(0)}d^ny\int_{B_R(0)}d^nx\,|f(x+y)-f(x)|\\
≤\delta_k\sup_{y\in B_{\delta_k(0)}}\int_{B_R(0)}d^nx\, |f(x+y)-f(x)|≤\delta_k \sup_{y\in B_{\delta_k}(0)}\|T_{y}(f)-f\|_{L^1}$$
Where $T_y$ is the translation of $y$. Now its clear that as $A_k\to\Bbb1$ you have that $\delta_k\to0$, as this is the maximal displacement $A_k$ does in the ball of radius $R$. Its also true that $\|T_y(f)-f\|$ to $0$, but we don't need that we just need that $T_y$ is an isometry and then the entire expression is smaller than $2\delta_k \|f\|_{L^1}$, which goes to zero. So the bound can be achieved.
