If f and g are simple functions such that $f \le g$ then the Lebesgue integral $\int_Xfd\mu \le \int_Xgd\mu$

Question

A common exercise I see in textbooks is the following:

If f and g are simple functions such that they are absolutely convergent then: $$f \le g \Rightarrow \int_Xfd \mu \le \int_Xgd \mu$$

This video gives a proof for the case of regular functions, and so I know that that proof would indeed follow for simple functions, but I was hoping someone could show me the proof when we aren't considering the general Lebesgue integral, but just the Lebesgue integral of simple functions. That is: $\int_Xfd \mu := \sum^\infty_{n=1}a_n\mu(f^{-1}(\{a_n\})$.

Textbooks always leave this as an exercise for the reader and so I don't think it's too hard, but for whatever reason I'm having trouble!

Answer 1

Simple functions have a finite amount of values, so if $f$ is a simple function then $\text{im}(f)=\{a_1,\ldots,a_n\}$. If $g$ is another simple function then $\text{im}(g)=\{b_1,\ldots,b_m\}$.

We know $X=\cup_{l=1}^m g^{-1}(\{b_l\})$, since every $x\in X$ gives us one of those values as its image under $g$; and this union is of pairwise disjoint sets, since no $x\in X$ could have two different images. Then $f^{-1}(\{a_k\})=\cup_{l=1}^mf^{-1}(\{a_k\})\cap g^{-1}(\{b_l\})$ as a union of pairwise disjoint sets, so by $\sigma$-additivity of $\mu$ we have

$\mu(f^{-1}(\{a_k\}))=\sum_{l=1}^m\mu(f^{-1}(\{a_k\})\cap g^{-1}(\{b_l\}))$.

Similarly $\mu(g^{-1}(\{b_l\})=\sum_{k=1}^n\mu(f^{-1}(\{a_k\})\cap g^{-1}(\{b_l\}))$.

Therefore $\int_Xf\,\text{d}\mu=\sum_{k=1}^na_k\mu(f^{-1}(\{a_k\}))=\sum_{k=1}^na_k\sum_{l=1}^m\mu(f^{-1}(\{a_k\})\cap g^{-1}(\{b_l\}))=$

$=\sum_{k=1}^n\sum_{l=1}^ma_k\mu(f^{-1}(\{a_k\})\cap g^{-1}(\{b_l\}))=\sum_{l=1}^m\sum_{k=1}^na_k\mu(f^{-1}(\{a_k\})\cap g^{-1}(\{b_l\}))$.

For those $f^{-1}(\{a_k\})\cap g^{-1}(\{b_l\})$ not empty, every $x$ there is satisfies $f(x)=a_k$ and $g(x)=b_l$, so by hypothesis $a_k\le b_l\Rightarrow a_k\mu(f^{-1}(\{a_k\})\cap g^{-1}(\{b_l\}))\le b_l\mu(f^{-1}(\{a_k\})\cap g^{-1}(\{b_l\}))$.

For those $f^{-1}(\{a_k\})\cap g^{-1}(\{b_l\})$ empty we have $\mu(f^{-1}(\{a_k\})\cap g^{-1}(\{b_l\}))=0$ so $a_k\mu(f^{-1}(\{a_k\})\cap g^{-1}(\{b_l\}))=0=b_l\mu(f^{-1}(\{a_k\})\cap g^{-1}(\{b_l\}))$.

Now we can do $\int_X f\,\text{d}\mu=\sum_{l=1}^m\sum_{k=1}^na_k\mu(f^{-1}(\{a_k\})\cap g^{-1}(\{b_l\}))\le\sum_{l=1}^m\sum_{k=1}^nb_l\mu(f^{-1}(\{a_k\})\cap g^{-1}(\{b_l\}))=$

$=\sum_{l=1}^mb_l\sum_{k=1}^n\mu(f^{-1}(\{a_k\})\cap g^{-1}(\{b_l\}))=\sum_{l=1}^mb_l\mu(g^{-1}(\{b_l\}))=\int_X g\,\text{d}\mu$.