if f,g are both measurable functions whose range is $\mathbb{R}^+$ then $\int_X g d\varphi = \int_X fgd\mu$

Question

Im studying analisis on my own, actually reading the third edition of "Real and complex analisis" by Walter Rudin. Im stuck on this theorem.

Suppose that $f: X\to [0,\infty]$ is measurable, and $$\tag{1}\varphi(E)=\int_E fd\mu \ \ \ \ \ (E\ \text{measurable set)}$$ Then $\varphi$ is a measure and $$\tag{2}\int_X gd\varphi=\int_X fgd\mu$$ for all measurable $g:X\to [0,\infty]$

and as proof they have this:

If (2) holds for every simple and measurable functions then i understand why the general case is true. Take an $\{s_n\}$ of simple and measurable functions such that $0\leq s_n \leq g$ and $s_n \to g$, then $$\int_X gd\varphi=\lim_{n \to \infty}\int_X s_nd\varphi = \lim_{n \to \infty} \int_Xs_nfd\mu$$ Given that $f\geq 0$, then $s_nf\leq gf$ and $s_nf \to gf$, hence $$\lim_{n \to \infty} \int_Xs_nfd\mu = \int_Xgfd\mu$$ Tho, i cannot actually see why if $g=\chi_E$ then (1) shows that (2) holds, neither why this implies that (2) holds for every simple.
This my attempt: For $g$ measurable, take $\{t_n\}$ a family of simple and measurable functions such that $0\leq t_n =\sum_{i=1}^m \alpha_i^n\chi_{A_i^n}\leq g$ and $t_n \to g$. By definition, $$\int_Xgd\phi=\text{sup}\int_X t_nd\varphi = \text{sup}\sum_{i}\alpha_i^n\varphi(A_i^n)=\text{sup}\sum_{i}\alpha_i^n\int_{A_i^n}fd\mu=\text{sup}\sum_{i}\alpha_i^n\int_X \chi_{A_i^n}fd\mu$$ $$\tag{3}=\text{sup}\int_X\sum_{i}\alpha_i^n\chi_{A_i^n}fd\mu =\text{sup}\int_X t_nfd\mu = \int_X gfd\mu$$ Notice that i dont use the process suggested by Rudin and makes me wonder whether this even true. If it is true, can someone show me the proof suggested by Rudin. And if it is not true, please, point out where im mistaken.
Also, i have problem with understanding how actually u work with $\text{sup}\int_Xt_nd\mu=\underset{n\geq 1}{sup}\sum_{i=1}^{m(n)}\alpha_i^n\mu(A_i^n)$ and even if i wrote this correctly. Any clarification on that its well received.

Answer 1

Let $E \subseteq X$ be measurable. Then $$\int_X \chi_E\,\mathrm{d}\varphi = \int_E\mathrm{d}\varphi = \varphi(E) = \int_Ef\,\mathrm{d}\mu = \int_X\chi_Ef\,\mathrm{d}\mu.$$

That is, $$\int_X g \,\mathrm{d}\varphi = \int_X fg\,\mathrm{d}\mu$$ holds for any $g \in \{\chi_E : E\subseteq X \text{ is measurable}\}.$ From there, showing it's true for every simple $g$ is just by linearity of the integrals.

That is, you let $g = \sum_{k=1}^n c_k\chi_{E_k}$ where each $c_k \geq 0$ and each $E_k$ is measurable to find $$\begin{align*} \int_X g\,\mathrm{d}\varphi &= \int_X\sum_{k=1}^nc_k\chi_{E_k}\,\mathrm{d}\varphi \\ &= \sum_{k=1}^n c_k\int_X\chi_{E_k}\,\mathrm{d}\varphi \\ &= \sum_{k=1}^nc_k \int_X\chi_{E_k}f\,\mathrm{d}\mu \\ &= \int_X\sum_{k=1}^nc_k\chi_{E_k}f\,\mathrm{d}\mu \\ &= \int_Xfg\,\mathrm{d}\mu \end{align*}$$

Answer 2

You really should follow Rudin’s outline. It is clear and to the point.

• First, prove it for $g=\chi_E$ where $E$ is a measurable set. Then, we have \begin{align} \int_Xg\,d\phi=\int_X\chi_E\,d\phi:=\phi(E):=\int_Ef\,d\mu=\int_X\chi_Ef\,d\mu=\int_Xgf\,d\mu. \end{align} So, $(2)$ holds for $g=\chi_E$.
• Next, note that equation $(2)$ is linear with respect to $g$, meaning that the set of complex-valued functions $g$ which satisfy equation $(2)$ forms a vector space (a subspace of all complex-valued functions). This is simply by linearity of integrals. To be 100% clear: if $g_1,g_2$ satisfy the equation, then for any $c\in\Bbb{C}$ (actually at this stage we only need to worry about non-negative $g_1,g_2$ (i.e taking values in $[0,\infty]$) and $c\geq 0$) the same is true for $cg_1+g_2$. The proof is just linearity of integrals (I assume you already know why $\phi$ itself is a measure): \begin{align} \int_X(cg_1+g_2)\,d\phi&=c\int_Xg_1\,d\phi+\int_Xg_2\,d\phi\\ &=c\int_Xg_1f\,d\mu+\int_Xg_2f\,d\mu\\ &=\int(cg_1+g_2)f\,d\mu. \end{align} Btw these are the sorts of equal signs which Rudin simply expects you to do in your head.
• So by the previous two bullet points, you know that (2) holds for all simple functions, in particular for all non-negative simple functions. For an arbitrary non-negative measurable function $g$, you can find an increasing sequence of simple functions $\{s_n\}$ converging to $g$ pointwise. So, \begin{align} \int_Xg\,d\mu&=\lim_{n\to\infty}\int_Xs_n\,d\phi=\lim_{n\to\infty}\int_Xs_nf\,d\mu=\int_Xgf\,d\mu. \end{align} The first equal sign is by applying MCT since $s_n$ increases $g$ the second is because $s_n$ satisfies $(2)$, the third is because of MCT, since $s_nf$ increases to $gf$ (we’re using non-negativity of $f$ here once again).

Hence, (2) holds for all non-negative measurable $g$.