# Prove that if $0\leq f\leq h$ then $\int f \leq \int h$

I am trying to learn measure theory and unfortunately I am having great trouble with it. Rudin claims that this is trivial but I don't see it. I want to to first tackle the case where $$f, h$$ are simple. Assume that $$f\leq h$$ and $$f,h$$ are simple, then $$f=\sum_{1}^{m}a_i\chi_{A_i}$$ and $$h=\sum_{1}^{n}b_j\chi_{B_i}$$.

I have proven that when representing simple functions we can assume $$A_i$$ and $$B_i$$ are collections of disjoint sets. and I used this fact to show that if $$0\leq f$$ then $$0 \leq \int f$$

I was hoping to show that there exists a collection of sets $$C_i$$ such that for all $$A_i=\cup C_k$$ and $$B_j=\cup C_n$$ but to no avail. I also have noticed that for each $$a_i$$ there is some $$b_j$$ that $$a_i since for each particular $$x$$, $$f(x) and $$x$$ is only in one of the partitions.

Any hints would be appreciated.

• Try thinking about level sets
– kkc
Commented Sep 10, 2019 at 2:43
• Try thinking about level sets
– kkc
Commented Sep 10, 2019 at 2:43
• The thing stated in the title is false, for if $f$ is nonnegative, and $h = f$, then $\int f = \int h$, and hence is not strictly less than $\int h$. Commented Sep 10, 2019 at 2:45
• @JohnHughes Fixed, thank you! Commented Sep 10, 2019 at 3:13

Hint You are almost done.

Set $$g=h-f$$. Then $$g \geq 0$$ and hence, by what you proved $$\int g \geq 0$$.

If you are trying to avoid linearity then set $$f=\sum_{i=1}^{m}a_i\chi_{A_i},h=\sum_{j=1}^{n}b_j\chi_{B_j}$$ and define $$a_{m+1}=0, A_{m+1}=X - \cup_{i=1}^{m} A_i$$ and $$b_{m+1}=0, B_{n+1}=X - \cup_{j=1}^{n} B_j$$. Then $$f= \sum_{i=1}^{m+1} a_i \sum_{j=1}^{n+1} \chi_{A_i\cap B_j}= \sum_{i=1}^{m+1}\sum_{j=1}^{n+1} a_i \chi_{A_i\cap B_j}\\ h=\sum_{j=1}^{n+1}b_j\sum_{i=1}^{m+1}\chi_{A_i \cap B_j}=\sum_{j=1}^{n+1}\sum_{i=1}^{m+1} b_j\chi_{A_i \cap B_j}$$

• Thank you for the answer! I wanted to do that but the fact the integral is linear is proved later on. So I was hoping I could do that without using this trick. Commented Sep 10, 2019 at 3:13
• @Sorfosh See the second hint then. Commented Sep 10, 2019 at 3:39
• I do have a follow up question. Now to generalize to any measurable functions $f,h$. We can say, Let $f\leq h$. If there is a simple function $g$ such that $f\leq g \leq h$ then by definition $\int f \leq \int h$ as $\int g$ is an upper bound of $\int f$ but $\int g \leq \int h$. If no such simple function exists then $\int f=\int h$ as the set of approximating simple functions is the same. Is this correct? I define integral of non simple function as an upper bound of integrals of simple functions less than or equal to the funciton. Commented Sep 11, 2019 at 17:47
• @Sorfosh Be carefull that the second last line is a bit more subtle than it looks. Note that if $g$ is a simple function such that $g \leq h$, the fact that $g$ is not between $f$ and $h$ doesn't mean that $g \leq f$. It only means that $g \not\geq f$. Commented Sep 11, 2019 at 17:56
• How about this. Given any simple $g$, $g\leq f$ we know that $g\leq h$ thus if we have the set of integrals of simple functions less than h, we know it will contain all, (but perhaps more) integrals of simple functions less than f. If $A \subseteq B$ then $Sup(A)\leq Sup(B)$ and so the result follows Commented Sep 11, 2019 at 18:07

I believe Rudin uses the following definition: Let $$A$$ be a measurable set in $$\mathbb R$$ and $$\mathscr S$$ the collection of simple functions on $$A$$. Let $$f\ge 0$$ on $$A$$. Then

$$\int_A f:=\sup\{\int_A s:s\le f, s\in \mathscr S\}$$.

And then the claim $$is$$ trivial because if $$f\le h$$ then if $$s\in \mathscr S$$ and $$s\le f$$, then $$s\le h$$, so

$$\{s\le f, s\in \mathscr S\}\subseteq \{s\le h, s\in \mathscr S\}$$.