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Let $(X,M,\mu)$ be a measure space and $\phi$ a non-negative simple function.

Given $X_0\subset X$, $X_0\in M$ and $\mu(X\setminus X_0)=0$, how do I show that $ \int_{X_0} \phi~d\mu = \int_X \phi~d\mu.$

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    What about, $\int_{X_0\cup (X\setminus X_0)}\phi d\mu=\int_X \phi d\mu$.2012-02-04

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Using the formula $\displaystyle\int_{A\cup B} \phi~ d\mu = \int_A \phi ~d\mu + \int_B \phi~d\mu$ you have derived with $A=X_0$ and $B=X\setminus X_0$, we have $\int_{X} \phi~ d\mu = \int_{X_0} \phi ~d\mu + \int_{X\setminus X_0} \phi~d\mu.$ If we can prove that $\displaystyle\int_{X\setminus X_0} \phi~d\mu=0$, then we are done.

To prove this, note that $\phi$ is a nonnegative simple function, we have $\phi=\sum_{i=1}^na_i\textbf{1}_{A_i}$ for some measurable sets $A_i$ and $a_i\geq 0$. Then $0\leq\int_{X\setminus X_0} \phi~d\mu=\sum_{i=1}^na_i\int_{X\setminus X_0} \textbf{1}_{A_i}~d\mu=\sum_{i=1}^na_i\mu(A_i\cap(X\setminus X_0)).$ By assumption $\mu(X\setminus X_0)=0$, we have $\mu(A_i\cap(X\setminus X_0))\leq \mu(X\setminus X_0)=0$, which implies that last term in the above equality is zero. This proves that $\displaystyle\int_{X\setminus X_0} \phi~d\mu=0$.

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    @Bill: You are welcome.2012-02-04