F is integrable if and only if its positive and negative parts are

[Fredrik]

Homework Statement

Problem 2.6.3. in "Foundations of modern analysis", by Avner Friedman. Let f be a measurable function. Prove that f is integrable if and only if f⁺ and f^- are integrable, or if and only if |f| is integrable.

Homework Equations

Friedman defines "integrable" like this: An a.e. real-valued measurable function $f:X\to\overline{\mathbb R}$ is said to be integrable if there's a sequence $\langle f_n\rangle$ of integrable simple functions that's Cauchy in the mean and such that $f_n\to f$ a.e.

"Cauchy in the mean" means that it's a Cauchy sequence with respect to the L¹ norm (which hasn't been defined at this point). In other words, $\langle f_n\rangle$ is Cauchy in the mean if for all $\varepsilon>0$ there's an $N\in\mathbb Z^+$ such that for all $n,m\in\mathbb Z^+$,
$$\newcommand{\dmu}{\ \mathrm{d}\mu}
n,m\geq N\ \Rightarrow\ \int|f_n-f_m|\dmu<\varepsilon.$$

The Attempt at a Solution

I haven't looked at the "Cauchyness" yet. I'm still just focusing on the part about convergence a.e. If $f^+, f^-$ are integrable, it's easy to show that f is, because there are sequences $\langle (f^\pm)_n\rangle$ that converge a.e. to $f^\pm$, and we just need to find another sequence $\langle f_n\rangle$ that converges a.e. to f. All we have to do is to define $f_n=(f^+)_n-(f^-)_n$, and the result $f_n\to f$ a.e. follows from
$$|f_n-f|=|(f^+)_n-(f^-)_n-f^+-f^-|\leq|(f^+)_n-f^+|+|(f^-)_n-f^-|.$$ The proof that |f| is integrable is very similar. It's the converses of these results that I'm struggling with. In particular, how do I prove that if f is integrable, then its positive and negative parts are integrable too?

It actually looks impossible to me. At least if I start with the "obvious" idea. Let $\langle f_n\rangle$ be a sequence of integrable simple functions that's Cauchy in the mean and such that $f_n\to f$ a.e. Then break each $f_n$ into positive and negative parts, and define the sequences $\langle (f^\pm)_n\rangle$ by $(f^\pm)_n=(f_n)^\pm$. (I'll just write $f^\pm_n$ from now on).

Consider an x such that f(x)=0. We have $f^\pm(x)=0$, but I don't see a reason why we can't have something like $f^\pm_n\to\pm 1$.

On the other hand, if I assume that both f and |f| are integrable (instead of just f), I can prove that f⁺ and f^- are integrable with a simple triangle inequality argument.

 

[micromass]

So your problem is: If $f_n\rightarrow f$ a.e. with $f_n$ simple, then also $f_n^+\rightarrow f^+$ a.e.

This is what you want to prove right??

But $f^+=f\vee 0$. So what you actually need to prove is that if $x_n\rightarrow x$, then $x_n\vee 0\rightarrow x\vee 0$. That doesn't sound too complicated to prove.

Am I missing something?

 

[Fredrik]

Yes, that's what I want to prove, and no I don't think you missed anything. I kept getting confused by inequalities like
$$|f^+_n(x)-f^+(x)|\leq |f_n(x)-f(x)|+|f^-_n(x)-f^-(x)|.$$ I see now that when x is such that f(x)>0, I can make the second term =0, not just <ε, by choosing ε<f(x) and N such that n≥N implies $|f_n(x)-f(x)|<\varepsilon$. So it looks like what I missed is that I have to choose ε small enough. Thanks for helping me figure that out.