Integral with respect to convolution of measure

Question

Let $E$ be a $\mathbb R$-Banach space, $$\theta_n:E^n\to E\;,\;\;\;x\mapsto\sum_{i=1}^nx_i$$ for $n\in\mathbb N$ and $\lambda$ be a measure on $\mathcal B(E)$. Remember that the $n$-fold convolution of $\mu$ with itself is given by the pushforard measure $$\lambda^{\ast n}:=\theta_n\left(\lambda^{\otimes n}\right)$$ for $n\in\mathbb N_0$; where $\lambda^{\ast0}:=\delta_0$.

Assume $\lambda$ is translation invariant; i.e. $$\lambda(B-x)=\lambda(B)\;\;\;\text{for all }B\in\mathcal B(E)\tag1.$$

Let $F$ be a $\mathbb R$-Banach space, $f\in\mathcal L^1(\lambda;F)$ and $n\in\mathbb N$. I would like to derive a formula for $$\lambda^{\ast n}f=\int f\:{\rm d}\lambda^{\ast n}.$$

If $f=\sum_{i=1}^kf_i1_{B_i}$ for some $k\in\mathbb N$ and disjoint $B_1,\ldots,B_k\in\mathcal B(E)$, then $$\lambda^{\ast n}(B_i)=\lambda(E)^{n-1}\lambda(B_i)\tag2$$ for all $i\in\{1,\ldots,k\}$ and hence $$\lambda^{\ast n}f=\lambda(E)^{n-1}\lambda f\tag2.$$

However, if we take $f=\operatorname{id}_E$, then we easily see (even without assuming $(1)$) $$\lambda^{\ast n}f=\sum_{i=1}^n\int x_i\:\lambda^{\otimes n}({\rm d}x)=n\lambda(E)^{n-1}\int x\:\lambda({\rm d}x)\tag3.$$ So, it seems like my formula $(2)$ is off by a factor of $n$.

But why? Is $(2)$ wrong? It should be correct, since \begin{equation}\begin{split}\lambda^{\otimes2}(B)&=\lambda^{\otimes 2}(\{x\in E^2:x_1+x_2\in B\})\\&=\int\lambda({\rm d}x_1)\lambda(B-x_1)=\lambda(E)\lambda(B)\end{split}\tag4\end{equation} for all $B\in\mathcal B(E)$.

So, what am I missing?

Answer 1

There is no contradiction, because every translation-invariant, nonzero measure on a Banach space has $\lambda(E)=\infty$.

Proof: Suppose $\lambda$ is nonzero and translation invariant. We know $\lambda(A)>0$ for some set $A$. Letting $B_r$ be the ball of radius $r$ centered at the origin, choose some $r$ for which $\lambda(A\cap B_r)>0$ (which exists since $\lambda(A)=\lim_{r\to\infty} \lambda(A\cap B_r)$), and set $A_r=A\cap B_r$. Letting $u$ be any unit vector, the sets $$ A_r,\quad A_r+2ru,\quad A_r+4ru,\quad A_r+6ru,\;\;\dots $$ are pairwise disjoint with nonzero measure, implying the measure of the whole space is infinite.

Therefore, there is no problem with your two equations being off by a factor of $n$, since they both involve $\lambda(E)^{n-1}=\infty$ and $n\cdot \infty=\infty$.