# 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?

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