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I have a measure $\mu$ defined as follow: for every a measurable set $E \subset \mathbb{R}^3$

$$\mu(E)=\int_{E \cap B(0, 1)} \sqrt{x^2+y^2+z^2}d\lambda_3(x,y,z)$$

where $\lambda_3$ is the Lebesgue measure in $\mathbb{R}^3$.

I have to compute:

$$\int_{B(0,2)} (xy^2+y^2z)d\mu(x,y,z)$$.

Is it true that:

$$\int_{B(0,2)} (xy^2+y^2z)d\mu(x,y,z)=\int_{ B(0, 1)} (xy^2+y^2z)\sqrt{x^2+y^2+z^2}d\lambda_3(x,y,z)$$?

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Yes. In general if $\mu(E) = \int_E f \, d\nu$ (on some appropriate measure space $X$ with appropriate measurability hypotheses regarding $f$ and $E$) it follows more-or-less straight from the definition that $$\int_X s \, d\mu = \int_X sf \, d\nu$$ for any simple function $s$, and via the usual limiting processes that $g \in L^1(\mu)$ whenever $g$ is measurable and $gf \in L^1(\nu)$, and moreover that $$\int_X g \, d\mu = \int_X fg \, d\nu$$

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