Sheldon Axler Real Analysis Exercise 16 Section 3A This is an exercise I've been working a whole day. I wrote a solution involving Zorn's lemma, but I'm not really satisfied using this non-standard result. I'm wondering if there is an easier solution that gets around this lemma.
Suppose $\mathcal{S}\subset\mathcal{T}$ are $\sigma$-algebras on $X$, $\mu_1$ and $\mu_2$ are measures on $(X,\mathcal{S})$ and $(X,\mathcal{T})$, respectively, and $\mu_1(E)=\mu_2(E)$ $\forall \,E\in\mathcal{S}$. Prove that if $f:X\to[0,\infty]$ is $\mathcal{S}$-measurable, then $\int fd\mu_1=\int fd\mu_2$.
Edit: $(X,\mathcal{S})$ and $(X,\mathcal{T})$ are generic measurable spaces, and my definition of integral is $\int fd\mu=\sup\{\sum_{A_i\in P}\mu(A_i)\inf_{A_i}(f)|P\text{ is a partition of } X\}=\sup\{\sum_i c_i\mu(A_i)|\sum_i c_i\chi_{A_i}\leq f\}$.
Here the $c_i$ can be chosen arbitrarily from $\mathbb{R}$, so I didn't expect to partition the range of $f$.
My reasoning:
Cleary $\int fd\mu_1\leq\int fd\mu_2$ since $\mathcal{S}\subset\mathcal{T}$, so we need the inequality on the other direction.
No matter how our strategy is, we need to say that for any partition of $\{F_i\}\subset2^\mathcal{T}$, we can find another partition $\{E_i\}\subset2^\mathcal{T}$ that approximates the integral better. The choice of $\{E_i\}$ clearly will depend on $\{F_i\}$, so we kind of need to find a refinement using only sets from $\mathcal{S}$. The problem is that if some $F_i\in2^\mathcal{T}-2^\mathcal{S}$ then there will be some elements that "goes out of" $F_i$ and the refinement will not work. However we can go around this if we can find a $F_i\subset E_i$ such that $f$ has the same range on them, and here comes Zorn's lemma.
Proof:
By Zorn's lemma, there exists a minimal set $E_i\in\mathcal{S}$ such that $F_i\subset E_i$. Then $$f(F_i)\subset f(E_i)$$ and $$F_i\subset f^{-1}(f(F_i))\cap E_i\subset f^{-1}(f(E_i))\cap E_i=E_i$$. But $f$ is $\mathcal{S}$-measurable, so $f^{-1}(f(F_i))\cap E_i\in\mathcal{S}$, and by minimality $f^{-1}(f(F_i))\cap E_i=E_i\Longleftrightarrow E_i\subset f^{-1}(f(F_i))$. Finally $$f(E_i)\subset f(f^{-1}(f(F_i)))\subset f(F_i)\subset f(E_i)\Longrightarrow f(F_i)=f(E_i)$$.
Now we can consider the $m$ subsets $\tilde{E_j}$ generated by intersection of the $n$ $E_i$s: These form a partition. Then $$\sum_{F_i}\mu_2(F_i)\inf_{F_i}(f)=\sum_{F_i}\sum_{\tilde{E_j}}\mu_2(F_i\cap\tilde{E_j})\inf_{E_i}(f)\leq\sum_{\tilde{E_j}}\sum_{F_i}\mu_2(F_i\cap\tilde{E_j})\inf_{\tilde{E_i}}(f)=\sum_{\tilde{E_j}}\mu_1(\tilde{E_j})\inf_{\tilde{E_i}}(f)$$ as desired.
 A: If $f$ is an $\mathcal{S}$-measurable indicator function, the claim is true by assumption. If $f$ is an $\mathcal{S}$-measurable nonnegative simple function, then the claim is true by linearity. Now suppose $f$ is an arbitrary $[0, \infty]$-valued $\mathcal{S}$-measurable function. Let $(f_j)$ be a sequence of $\mathcal{S}$-measurable nonnegative simple functions increasing pointwise to $f$. We have $\int f_j \,d\mu_1 = \int f_j \,d\mu_2$ for all $j$. Letting $j \to \infty$ and using the monotone convergence theorem yields $\int f \,d\mu_1 = \int f \,d\mu_2$.
A: I do not agree that this is obvious: It really depends on the definition you use for $\int f d\mu$.  If you chop up the range it falls out immediately, but I suspect you are using a definition of $\int f d\mu$ as the supremum integral over simple functions $g$ such that $g \leq f$, which means you are chopping up the domain.
You can use this fact (which also helps to understand the relationship between chopping up the range or the domain):
Claim: If $g:X\rightarrow \mathbb{R}$ is a function that takes at most $n$ values $v_1<v_2< ...< v_n$, and if $g(x)\leq f(x)$ for all $x\in X$ and for some function $f:X\rightarrow (-\infty, \infty]$, then
$$ g(x) \leq h(x) \equiv v_n1_{\{f(x) \in [v_n, \infty]\}} + \sum_{i=1}^{n-1} v_i 1_{\{f(x) \in [v_i,v_{i+1})\}}  \quad \forall x \in X$$
Proof: Fix $x \in X$. Then $g(x) = v_j$ for some $j \in \{1, ..., n\}$, and so $f(x)\geq v_j$, which means either $f(x) \in [v_n, \infty]$ or $f(x) \in [v_i, v_{i+1})$ for some $i\geq j$, and so  $h(x) \geq v_j$. $\Box$
A: You are overthinking this. No choice or Zorn’s Lemma is required for this proof.
Let us say we find some disjoint finite collection of pairwise disjoint sets $T_1, T_2, \ldots, T_n \in \mathcal{T}$, together with numbers $0 < a_1 < a_2 < \ldots < a_n < \infty$, such that for all $i$, for all $x \in T_i$, $a_i < f(x_i)$. For convenience, define $a_0 = 0$ and $a_{n + 1} = \infty$.
Then define $E_i = f^{—1}((a_i, a_{i + 1}])$. I claim that $\sum\limits_{i = 1}^n \mu_2(T_i) a_i \leq \sum\limits_{i = 1}^n \mu_2(E_i) a_i$.
To prove this, note that for all $i$, we have $T_i \subseteq f^{-1}((a_i, \infty]) = f^{-1}(\bigcup\limits_{j = i}^n (a_j, a_{j + 1}]) = \bigcup\limits_{j = i}^n f^{-1}((a_j, a_{j + 1}]) = \bigcup\limits_{j = i}^n E_j$. Then we see that $T_i = (\bigcup\limits_{j = i}^n E_j) \cap T_i = \bigcup\limits_{j = i}^n E_j \cap T_i$. Then we see that $\mu_2(T_i) = \mu_2(\bigcup\limits_{j = i}^n E_j \cap T_i) = \sum\limits_{j = i}^n \mu_2(E_j \cap T_i)$. In particular, then, we have $\mu_2(T_i) a_i = \sum\limits_{j = i}^n \mu_2(E_j \cap T_i) a_i$.
Therefore, we have
$\begin{equation}
\begin{split}
\sum\limits_{i = 1}^n \mu_2(T_i) a_i 
&= \sum\limits_{i = 1}^n \sum\limits_{j = i}^n \mu_2(E_j \cap T_i) a_i \\
&= \sum\limits_{1 \leq i \leq j \leq n} \mu_2(E_j \cap T_i) a_i \\
&\leq \sum\limits_{1 \leq i \leq j \leq n} \mu_2(E_j \cap T_i) a_j \\
&= \sum\limits_{j = 1}^n \sum\limits_{i = 1}^j \mu_2(E_j \cap T_i) a_j \\
&= \sum\limits_{j = 1}^n (\sum\limits_{i = 1}^j \mu_2(E_j \cap T_i)) a_j \\
&= \sum\limits_{j = 1}^n \mu_2(\bigcup\limits_{i = 1}^j E_j \cap T_i) a_j \\
&= \sum\limits_{j = 1}^n \mu_2(E_j \cap \bigcup\limits_{i = 1}^j T_i) a_j \leq \sum\limits_{j = 1}^n \mu_2(E_j) a_j
\end{split}
\end{equation}$
which completes the proof of my claim.
Now note that for all $i$, we have $E_i \in \mathcal{S}$ by measurability of $f$.
This result is therefore sufficient to establish that $\int f d\mu_2 \leq \int f d\mu_1$.
