measure theory – If $left.lambdaright|_r$ is the restriction of $λ$ to the ball of radius $r$, can we show $left.λright|_1astleft.λright|_1=left.λright|_2$?

This shouldn’t be to difficult, but I struggle to obtain the following result:

Let $lambda$ be a finite nonnegative measure on the Borel $sigma$-algebra $mathcal B(E)$ of a normed $mathbb R$-vector space $E$, $$overline B_r:=left{xin E:left|xright|_Ele rright}$$ and $$left.lambdaright|_r(A):=lambdaleft(Acapoverline B_rright);;;text{for }Ainmathcal B(E)$$ for $r>0$.

Are we able to show $left.lambdaright|_1astleft.lambdaright|_1=left.lambdaright|_2$?

As usual, if $mu_i$ is a finite signed measure on $mathcal B(E)$, then $$(mu_1astmu_2)(A):=(mu_1astmu_2)left(left{(x,y)in E^2:x+yin Aright}right)=intmu_1({rm d}x)mu_2(A-x)tag1$$ for $Ainmathcal B(E)$.

Intuitively, we may clearly write $$overline B_2=overline B_1cupbigcup_{xinoverline B_1}left{x+y:yinoverline B_1right}tag2,$$ which should somehow immediately yield the result.

So, how can we prove the desired claim?