Detail in standard measure theory I cannot seem to obtain

Question

There is a standard result in measure/integration theory which I just cannot seem to obtain.

If $f \colon X \to \mathbb{C}$ is measurable ($X$ is any measurable space), there exist simple measurable functions $\phi_k \colon X \to \mathbb{C}$ such that $\phi_k \to f$ pointwise and all $|\phi_k| \le |f|$. This is fine.

However, it is often claimed that the $\phi_k$ can be taken such that also $|\phi_k| \le |\phi_{k+1}|$ for all $k$. (See for instance Folland's Real Analysis, Proposition 2.10.) This I cannot seem to obtain, at least not with the simplicitly for which it is claimed to follow.

It is typically stated that this can be obtained as follows (e.g., Folland): Write $f = u + iv = u^+ - u^- + i(v^+ - v^-)$, the standard decomposition into positive and negative parts $u^\pm$ and $v^\pm$ for $u = \Re f$ and $v = \Im f$. Pick simple measurable functions $s_k^\pm$ and $t_k^\pm$ such that $0 \le s_k^\pm \uparrow u^\pm$ and $0 \le t_k^\pm \uparrow v^\pm$. Put $$\phi_k := s_k^+ - s_k^- + i(t_k^+ - t_k^-).$$ The $\phi_k$ are simple measurable functions $X \to \mathbb{C}$, and clearly $\phi_k \to f$ pointwise. All good.

At this point it is typically claimed that also $|\phi_k| \le |\phi_{k+1}|$ for all $k$, but how is this obtained in this exact setup? It appears standard wisdom is that $|\phi_k| = s_k^+ + s_k^- + t_k^+ + t_k^-$, from which it would follow easily indeed, but I am skeptical of this identity.

Suppose for the moment that $f$ is real (but please address the complex case), so that $v = 0$, hence $t_k^\pm = 0$. Then $\phi_k = s_k^+ - s_k^-$, a difference of two positive functions. From this follows $$|\phi_k| = \phi_k^+ + \phi_k^- \le s_k^+ + s_k^-,$$ but it is easy to cook up positive functions $s_k^\pm$ (below) for which the above inequality is strict, i.e., $|\phi_k| < s_k^+ + s_k^-$. Hence it cannot be the case that $|\phi_k| = s_k^+ + s_k^-$ in general; it appears to me a common pitfall is to confuse oneself from the identity $\phi_k = s_k^+ - s_k^-$ to believe that $s_k^\pm$ must be the positive and negative parts of $\phi_k$ (e.g., last computation here), but this is not true unless there is a trick involved. What is this trick?

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Example of strict inequality in a positive decomposition: Take $u := 1_{(-\infty, 1]}$ and $v := 1_{[-1, \infty)}$ on the real line. Take $f := u - v$. Then in general $f^+ \le u$ and $f^- \le v$, and in this case there holds $|f| = f^+ + f^- < u + v$.

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I suppose I could salvage the proposition by working out at a detailed level, but the proposition is claimed to follow leisurely from the approximation of positive measurables, so surely there is a trick here I am missing?

Answer 1

You need to also use the fact that these approximations are subordinate to the positive and negative parts of $u,v$. Here’s a more precise version of what I mean:

Lemma

Let $u:X\to[-\infty,\infty]$ be a given function and $u^+,u^-$ its positive, negative parts. Suppose $0\leq s_1\leq u^+$ and $0\leq s_2\leq u^-$ are given functions. Then, the subtraction $s:=s_1-s_2$ is a well-defined function $X\to [-\infty,\infty]$, with $s^+=s_1$ and $s^-=s_2$.

We cannot have $s_1(x)=s_2(x)=\infty$, because that would mean $u^+(x)=u^-(x)=\infty$, which is absurd (in fact if $u^+(x)>0$ that automatically implies $u^-(x)=0$, and vice-versa); hence the subtraction $s:=s_1-s_2$ is well-defined.

• Suppose $s(x)=0$. Then, $s_1(x)=s_2(x)$. If this common value is strictly positive, then $u^+(x)>0$ and $u^-(x)>0$, which as I said above is impossible by definition. Hence, $s_1(x)=s_2(x)=0$. So, $s^+(x):=\max(s(x),0)=\max(0,0)=0=s_1(x)$, and similarly, $s^-(x)=s_2(x)$.
• Suppose $s(x)>0$. Then, $s_1(x)>s_2(x)\geq 0$, which implies $u^+(x)>0$ and so $u^-(x)=0$ and hence $s_2(x)=0$. So, $s^+(x)=\max(s(x),0)=\max(s_1(x),0)=s_1(x)$, since it is positive; also we have $s^-(x)=-\min(s(x),0)=-\min(s_1(x),0)=-0=0=s_2(x)$.
• Suppose $s(x)<0$. Then, $s_2(x)>s_1(x)\geq 0$, and so $u^-(x)>0$, which implies $u^+(x)=0$, and thus $s_1(x)=0$, and so $s(x)=-s_2(x)$. It follows that $s^+(x)=0=s_1(x)$ and $s^-(x)=s_2(x)$.

This covers all the cases, so indeed $s^+=s_1$ and $s^-=s_2$.

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Now, let $f:X\to\Bbb{C}$ be a given function, $f=u+iv=(u^+-u^-)+i(v^+-v^-)$ be its decomposition. Suppose we take any functions $0\leq s^+\leq u^+, 0\leq s^-\leq u^-, 0\leq t^+\leq v^+, 0\leq t^-\leq v^-$, and define $s=s^+-s^-$ and $t=t^+-t^-$ and $\phi=s+it$. By the previous lemma, these really are the positive and negative parts, so it’s not an abuse of notation. So, we have \begin{align} |\phi|&=\sqrt{s^2+t^2}=\sqrt{|s|^2+|t|^2}=\sqrt{(s^++s^-)^2+(t^++t^-)^2}. \end{align} Now, if you put a subscript $k$ everywhere, then obviously we have the increasing nature of everything.