Proof of The Integral of an Indicator Function is the Measure of the Set

For a subset $E$ of the real line, the preimage $\mathbf{1}_A^{-1}(E)$ is $X$ if $E$ contains both $0$ and $1$, is $A$ if $E$ contains $1$ but not $0$, is $X\setminus A$ if $E$ contains $0$ but not $1$, and is $\varnothing$ otherwise; the set $X\setminus A$ lies in $\mathcal{F}$ because a $\sigma$-algebra is closed under complements. All four sets therefore lie in $\mathcal{F}$, so $\mathbf{1}_A$ is measurable in the sense required by Simple Function and Its Integral; it takes at most the two values $0$ and $1$ and is nonnegative, so it is a nonnegative simple function.

It remains to compute its integral from the standard representation of Simple Function and Its Integral, distinguishing the degenerate cases.

If $X=\varnothing$ then $A=\varnothing$, the function $\mathbf{1}_A$ takes no values, its standard representation is an empty sum, its integral is the empty sum $0$, and $\mu(A)=\mu(\varnothing)=0$.

If $X\ne\varnothing$ and $A=X$, the only value is $1$, with $\mathbf{1}_A^{-1}(\{1\})=X$, so the integral is $1\cdot\mu(X)=\mu(A)$.

If $X\ne\varnothing$ and $A=\varnothing$, the only value is $0$, with $\mathbf{1}_A^{-1}(\{0\})=X$, so the integral is $0\cdot\mu(X)=0=\mu(A)$, using the convention $0\cdot\infty=0$ of Measure, Measure Space, and Probability Measure.

In the remaining case the distinct values are $1$ and $0$, with $\mathbf{1}_A^{-1}(\{1\})=A$ and $\mathbf{1}_A^{-1}(\{0\})=X\setminus A$, so the integral is

again by the convention $0\cdot\infty=0$.

In every case the integral of $\mathbf{1}_A$ as a simple function is $\mu(A)$, and Lebesgue Integral of a Nonnegative Measurable Function records that for a nonnegative simple function that integral agrees with the integral defined there.