# Integral of borel function wrt to measure

Let $$(\Omega,\mathscr{F},v)$$ be some measure space. Consider the non-negative simple function $$l:\Omega \rightarrow R$$ given by:

$$l(\omega)=\sum _{i=1}^k a_i I_{A_{i}}(\omega)$$

Where $$a_i\in R_+, A_i\in\mathscr{F}$$, and $$I_{A_{i}}$$ is the indicator function for $$A_i$$

Definition 1: Integral of a non-negative simple function $$l$$ wrt to a measure $$v$$ is given by:

$$\int l~dv = \sum_{i=1}^k a_i v(A_i)$$

Definition 2: Integral of some non-negative Borel measurable function $$f:\Omega\rightarrow R$$ with respect to measure $$v$$ is given by:

$$\int f~dv= \sup\left\{\int l~dv: l \in S_f \right\}$$

Where $$S_f$$ is a collection of non-negative simple functions satisfying $$l(\omega) \leq f(\omega)$$ $$\forall\omega\in\Omega$$.

Consider some non-negative simple function $$l'$$. I am aware all simple functions as defined above are Borel measurable. So how do I reconcile definition 1 and 2? That is, how do I prove that:

$$\sup\left\{\int l~dv: l \in S_{l'} \right\} = \sum_{i=1}^{k'} a_i' v(A_i')$$

Where $$l'(\omega)=\sum _{i=1}^{k'} {a_i'} I_{A_{i}'}(\omega), a_i'\in R_+, A_i'\in\mathscr{F}$$ and $$S_{l'}$$ is the collection of non-negative simple functions satisfying $$l(\omega)\leq l'(\omega)$$ $$\forall \omega\in \Omega$$

Hint: Supose you haev $$l=\sum _{i=1}^k a_i I_{A_{i}}(\omega)$$ and $$l=\sum _{i=1}^m b_i I_{B_{i}}(\omega)$$ with $$l \leq l'$$. Consider the collection $$\{A_i\cap B_j : 1 \leq i \leq k, 1 \leq j \leq m\}$$. Write this as $$\{C_1,C_2,...,C_n\}$$. Then we can write $$l(\omega)=\sum _{i=1}^n c_i I_{C_{i}}(\omega)$$ and $$l'(\omega)=\sum _{i=1}^n d_i I_{C_{i}}(\omega)$$ for some $$c_i$$'s and $$d_i$$'s. We now have $$c_i \leq d_i$$ for $$1 \leq i \leq n$$ and this gives $$\sum _{i=1}^n c_i v(C_{i}) \leq \sum _{i=1}^n d_i v(C_{i})$$. If $$f=l'$$ is a simple function then supremum in the second definition is attained when $$l=l'$$.

Definition $$1$$ as stated in your answer should at least be accompanied with a proof that it is well-defined, i.e. that $$\sum_{i=1}^ka_i1_{A_i}=\sum_{j=1}^mb_j1_{B_j}$$ implies that $$\sum_{i=1}^ka_i\nu(A_i)=\sum_{j=1}^mb_j\nu(B_j)$$.

Let us go for a definition with the same strength that avoids that:

Let $$l$$ be a measurable function with finite range $$\{a_1,\dots,a_k\}\subseteq\mathbb [0,\infty)$$ where the $$a_i$$ are demanded to be distinct.

Then according to our new definition $$1$$: $$\int ld\nu:=\sum_{i=1}^ka_i\nu(A_i)\text{ where }A_i=l^{-1}(\{a_i\})\text{ for }i=1,\dots,k$$

If similarly $$l'$$ is a measurable function with finite range $$\{b_1,\dots,b_m\}\subseteq\mathbb [0,\infty)$$ where the $$b_j$$ are demanded to be distinct and moreover $$l'(\omega)\leq l(\omega)$$ for every $$\omega\in\Omega$$ then it is essential to prove in this context that $$\int l'd\nu\leq\int ld\nu$$.

For this observe that:$$\int l'd\nu=\sum_{j=1}^{m}b_{j}\nu\left(B_{j}\right)=\sum_{j=1}^{m}\sum_{i=1}^{k}b_{j}\nu\left(A_{i}\cap B_{j}\right)$$$$\leq\sum_{j=1}^{m}\sum_{i=1}^{k}a_{i}\nu\left(A_{i}\cap B_{j}\right)=\sum_{i=1}^{k}\sum_{j=1}^{m}a_{i}\nu\left(A_{i}\cap B_{j}\right)=\sum_{i=1}^{k}a_{i}\nu\left(A_{i}\right)=\int ld\nu$$

Here $$B_j$$ denotes the inverse image of $$\{b_j\}$$ w.r.t $$l'$$ and the inequality is justified by the fact that $$b_{j}=l'\left(\omega\right)\leq l\left(\omega\right)=a_{i}$$ whenever $$\omega\in A_{i}\cap B_{j}$$. If no such $$\omega$$ exists then $$A_{i}\cap B_{j}=\varnothing$$ and consequently $$\nu\left(A_{i}\cap B_{j}\right)=0$$, and in that case the corresponding terms on both sides fall out.

Now it can easily be proved that the two definitions reconcile.