Lebesgue Integral of the indicator (characteristic) function.

Question

For $(X, \mathcal A, \mu)$, $x \in X$ and $A \in \mathcal A$, is the following accurate? $$\int_X 1_A(x) \mu(dx) = \int_A \mu(dx) = \mu(A)$$

I'm not sure if my instincts are correct here. Thanks for any clarity.

Answer 1

Definition (Lebesgue Integral). Let $(X, \mathcal A, \mu)$ be a measure space and let $E \in \mathcal A$ be a measurable set. Let $s : X \to [0,\infty)$ be a measurable step function of the form $$s=\sum_{i=1}^{n}\alpha_i\mathbf{1}_{A_i},$$ where $α_i ∈ [0,∞)$ and $A_i ∈ \mathcal A$ for $i = 1, \dots , n$. The (Lebesgue) integral of $s$ over $E$ is the number $\int_E s\,\mathrm d\mu ∈ [0,∞]$ defined by $$\int_E s\,\mathrm d\mu=\sum_{i=1}^n\alpha_i\mu(E\cap A_i).$$

Define $A_1=A$ for the sake of noation. Let $A\subseteq E$. The indicator function $\mathbf 1_A$ is simple. We may write it as $$s= \sum_{i=1}^{n}\alpha_i\mathbf{1}_{A_i},$$ where $\alpha_1=1$ and $\alpha_i=0$ for $i>1$. Our integral is therefore equal to $$\int_E \mathbf 1_A\,\mathrm d\mu=\mu(E\cap A).$$ Since $A\subseteq E$, we have $\mu(A)$, as desired.