Decompose a simple function into the sum of two simple functions

Let $(\Omega, \mathscr F)$ be a measurable space and let $f,g : \Omega \to \mathbb R$ be non-negative Borel functions. Suppose that $\psi$ is a non-negative simple function, i.e. $\psi = \sum_{i=1}^n a_i \mathbf 1_{A_i}$ for some partition $\{A_i\}$ of $\Omega$ and some non-negative real numbers $\{a_i\}$, and that $\forall \omega \in \Omega$ $0 \leq \psi(\omega) \leq f(\omega) + g(\omega)$. By $\mathbf 1_{A}$ I mean the indicator function for the set $A$.

I want to decompose $\psi$ into the sum of two simple functions $\varphi$ and $\varphi'$ where $0 \leq \varphi \leq f$ and $0 \leq \varphi' \leq g$.

I believe this to be possible but I'm finding it difficult to prove. One example that I've been thinking about is taking $\Omega = [0, \pi/2]$, $f(x) = \sin(x) + 1$, and $g(x) = -\sin(x) + 1$. This means that $f + g = 2$, so I can take $\psi(x) = 1.99 \cdot \mathbf 1_{[0,\pi/2]}$. I can split this $\psi$ up, but I end up needing $\varphi$ and $\varphi'$ to be Riemann sum-like functions involving a large number of steps. This suggests that a constructive proof will be very difficult since $f + g$ and $\psi$ may be very simple, yet $f$ and $g$ separately can be extremely complicated.

I've considered trying to show that we can find $\varphi$ and $\varphi'$ such that their sum is arbitrarily close to $\psi$, but that would require some kind of norm on my simple functions and I haven't introduced that yet. I haven't even mentioned a measure. I'm hoping there's a proof that doesn't require introducing new machinery like that.

Update

@Hagen von Eitzen has pointed out that this is not generally true if $\psi \leq f + g$ is allowed, so I will require $\psi < f + g$.

1 Answer

Let $\Omega = (0,1]$, $f(x) = \frac{3x}{2}$, $g(x) = 1-\frac{x}{2}$, and $\psi \equiv 1$. Then $\psi < f+g$ and we cannot decompose $\psi$ as described in the question. If $\psi = \phi+\phi'$, where $\phi \le f$, then one term of $\phi$ must be $0\chi_E$ since $f$ gets arbitrarily small. So, $\phi'$ must have a term $\chi_E$, but $g$ is always less than $1$, so we won't have $\phi' \le g$.

Here are some of my thoughts on the problem (which explain the motivation for the counterexample above and might be helpful in case you modify the problem). It's not too hard to see we can assume $\psi \equiv \chi_\Omega$ and that $f+g > 1$. If such a decomposition $\psi = \phi+\phi'$ existed, then we can take the common refinement of the partitions of $\phi$ and $\phi'$ so that $\phi = \sum_{i=1}^n a_i\chi_{E_i}$, $\phi' = \sum_{i=1}^n (1-a_i)\chi_{E_i}$ where $\{E_1,\dots,E_n\}$ partitions $\Omega$. If $\{0 \le f \le \frac{1}{N}\}$ is non-empty, then some $a_i$ must be $0$, so $g$ must be $\ge 1$ on $E_i$. This need not happen, as my example above indicates. I do think this problem is weird in some sense. Basically, the only place where the measure structure is occurring is that $f$ and $g$ must be measurable, but the problem never has a.e. conditions. There's not enough structure here - $f$ and $g$ can be very wild while $f+g$ can be simple (no pun intended), as you explained. Having a simple function be dominated by $f$ is a pretty strong restriction..