Is Zorn's lemma applicable to this integration problem? Please consider the following problem and my proposed solution.
Suppose $(X,\mathcal{M},μ)$ is a measure space, and $\mathcal{F}$ is a family of non-negative integrable
functions on $X$ with the following properties:
(a) If $φ, ψ ∈F$, then $φ + ψ ∈ \mathcal{F}$.
(b) If $φ, ψ ∈F$, then $\max\{φ,ψ\}∈\mathcal{F}$.
(c) If $f$ is measurable, $f ≥ 0$ and $∫f dμ > 0$, then there is $φ ∈ \mathcal{F}$ such that $φ ≤ f$
and $∫φdμ > 0$.
Prove that if $f$ is non-negative and measurable then
$$
\int f dμ = \sup\left\{ \int \varphi \ \mathrm{d}\mu : \varphi \in \mathcal{F}, \varphi \le f\right\}$$
My Attempt
Assume that $\int f \ \mathrm{d}\mu > 0$. Then $\int f \ \mathrm{d}\mu$ is clearly an upper bound for the set above. We must prove that it is the least upper bound. Select $\varphi_1 \in \mathcal{F}$ such that $\varphi_1 \le f$ and $∫φ_1dμ > 0$. If $\varphi_1 = f$ a.e., then we are done, otherwise $\int (f - \varphi_1) \mathrm{d}\mu > 0$ and hence there is $\varphi_2 \in \mathcal{F}$ such that $\varphi_2 \le f - \varphi_1$, or $\varphi_1 + \varphi_2 \le f$ and $∫φ_2 \ dμ > 0$.
If $\varphi_1 + \varphi_2 \ne f$ a.e., then similarly we can find $\varphi_3$ such that $\varphi_1 + \varphi_2 + \varphi_3\le f$ and $∫φ_3 \ dμ > 0$. We may continue this process, if it does not end after $n$ steps to get
$$
\sum_{k=1}^n \varphi_n \le f.
$$
For each $n$, define $\Phi_n = \max\{\varphi_1,\dots,\varphi_n\}$. Then $\Phi_n \in \mathcal{F}$ using induction and property (b), and $\Phi_n \le \Phi_{n+1}$. Hence $\Phi_n \to \Phi$ for some measurable function $\Phi$, and by the monotone convergence theorem,
$$
\int \Phi_n \mathrm{d} \mu = \int \Phi \mathrm{d}\mu.
$$
If $\int \Phi \mathrm{d}\mu \ne \int f \mathrm{d}\mu$, then we may repeat this procedure yet again, starting instead with $f- \Phi$. If $\int f \mathrm{d} \mu = \infty$, then we can repeat this procdure indefinitely to find $\Phi$ so that $\int \Phi \mathrm{d}\mu$ exceeds any finite bound.
It seems to me that since we can always extend this procedure, we must eventually reach a maximal $\Phi$ and this $\Phi$ must be $f$, otherwise we may still extend. Now I am thinking of Zorn's lemma, but I am not exactly sure how it applies. I would like some clarification in that regard, or a different approach to the problem if possible. Thanks in advance.
 A: An argument along these lines can be made to work, though it seems rather awkward to force it into the framework of Zorn's lemma and it can be described more simply and naturally using transfinite recursion.  The main obstacle to making the idea work is that there is no reason to believe that your limiting function $\Phi$ is in $\mathcal{F}$.  So, you cannot simply repeat the procedure past that $\Phi$ if $\int \Phi<\int f$, since once you get past $\Phi$ the functions you are constructing are no longer elements of $\mathcal{F}$.  (Moreover, you should not expect the procedure to work in the way you describe, since as you suggest, it would end up giving an element of $\mathcal{F}$ that is actually equal to $f$ almost everywhere.  This is a much stronger conclusion than what you are asked to prove, and it is easy to come up with examples where it is not true.  For instance, $\mathcal{F}$ could consist of the simple functions, and it is certainly not true that any nonnegative measurable function has to be equal to a simple function almost everywhere.)
The trick to fix this is to replace $\Phi$ with $\Phi_n$ for sufficiently large $n$ in the "next step".  Specifically, suppose $\int \Phi<\int f$, and we take $\varphi\in \mathcal{F}$ with $\varphi\leq f-\Phi$ and $\int\varphi>0$.  We would like to now replace $\Phi$ with $\Phi+\varphi$, but that might not be in $\mathcal{F}$.  However, we can instead take $\Phi_n+\varphi$ for some large $n$.  If $n$ is sufficiently large, then $\int(\Phi_n+\varphi)$ will still greater than $\int\Phi$, and so we still have gotten closer to $\int f$ using elements of $\mathcal{F}$.
So, then, here is the full argument.  We iterate this construction by transfinite recursion, constructing a sequence of functions $(\varphi_\alpha)$ that are in $\mathcal{F}$ and below $f$.  At successor steps, if $\int\varphi_\alpha=\int f$ we are done, and otherwise we apply property (c) to $f-\varphi_\alpha$ to get $\varphi\in\mathcal{F}$ with $\varphi\leq f-\varphi_\alpha$ and $\int\varphi>0$, and we take $\varphi_{\alpha+1}=\varphi_\alpha+\varphi$.  At limit steps, we first take a limit of functions and then use the trick above to get an element of $\mathcal{F}$.  Specifically, suppose $\alpha$ is a countable limit ordinal and we have already defined $\varphi_\beta$ for each $\beta<\alpha$.  Enumerate the ordinals less than $\alpha$ as $(\beta_n)_{n\in\mathbb{N}}$ and let $\Phi_n=\max(\varphi_{\beta_0},\dots,\varphi_{\beta_{n-1}})$, and let $\Phi$ be the limit of this increasing sequence $(\Phi_n)$.  If $\int\Phi=\int f$, we are done, since each $\Phi_n$ is in $\mathcal{F}$ and $\int\Phi_n\to\int\Phi=\int f$.  If not, as in the previous paragraph, we can apply (c) to $f-\Phi$ and then take $n$ such that the resulting function $\Phi_n+\varphi$ still satisfies $\int(\Phi_n+\varphi)>\int\Phi$.  We then define $\varphi_\alpha$ to be this $\Phi_n+\varphi$.  Note that this $\varphi_\alpha$ satisfies $\int \varphi_\alpha>\int\Phi\geq\int\varphi_\beta$ for each $\beta<\alpha$.
If this transfinite recursion never terminates by reaching either a case where $\int\varphi_\alpha=\int f$ (in a successor step) or $\int\Phi=\int f$ (in a limit step), then it constructs a sequence $(\varphi_\alpha)$ over all countable ordinals $\alpha<\omega_1$ such that $\int\varphi_\alpha<\int\varphi_\beta$ whenever $\alpha<\beta$.  This is impossible, since there is no strictly increasing sequence of real numbers of length $\omega_1$.  So, we must eventually reach $\int\varphi_\alpha=\int f$ or $\int\Phi=\int f$ at some step, and the result is proved.
