Is this an alternative definition of the Lebesgue integral?

I am a complete beginner in measure theory and have only been reading Tao's notes on them for the past couple of days. Therefore what I say might be a triviality or just plain wrong.

Anyway, Tao covers the Jordan measure, followed by the Lebesgue measure followed by the Lebesgue integral. His definition of the Lebesgue integral is stepwise, first defining it for simple functions and then approximating functions using these simple functions.

This approach is a little complicated, for instance you have to show that the decomposition into simple functions is irrelevant and so on. However, given the definition of a Lebesgue measure, would the following also be a definition of the Lebesgue integral:

For a function $f: \mathbb R^n \to \mathbb R$, define: $$\int fd\mu = \mu(\overline\Gamma_+) - \mu(\overline\Gamma_-)$$ where: $$\overline\Gamma_+ = \{(x,y) \in \mathbb R^n\times\mathbb R : y \in [0,f(x)], f(x) \geq 0\}$$ and similarly: $$\overline\Gamma_- = \{(x,y) \in \mathbb R^n\times\mathbb R : y \in [0,f(x)], f(x) \leq 0\}$$

We will call $f$ Lebesgue integral if $\overline\Gamma$ is Lebesgue measurable. If $\mu$ is the Jordan measure, then it is not hard to see that this is simply the usual definition of the Riemann integral and is in fact, quite close to how we normally think about it.

Is there any reason not to use this definition?

• As you write it, this fails if $f(x)<0$ on more than a set of measure zero. But for positive $f$, this does serve as a definition of Lebesgue integral. Could you prove the standard convergence results, such a monotone and dominated convergence from it? – Lord Shark the Unknown Mar 2 '18 at 20:00
• @LordSharktheUnknown ah, I forgot to put in the negative sign for the negative portions. I guess with that in, it does indeed work in general? Edit: I fixed the sign issue. I will try and prove the convergence results. – Asvin Mar 2 '18 at 20:09
• My suspicion is that you can define the integral of $f$ in terms of its graph but you need to use simple functions to compute that volume and indeed, to prove anything meaningful about $\int f$. Also note that this requires you to look at $\mu \otimes \lambda$ where $\mu$ is your Borel measure on $\mathbf R^n$ and $\lambda$ is the Lebesgue measure on $\mathbf R^n$. Just having a measure on $\mathbf R^n$ doesn't give you, a priori, a measure on $\mathbf R^{n + 1}$. But now if you do this, you need to know about product measures. – Trevor Gunn Mar 2 '18 at 20:15
• @TrevorGunn: Product measures are okay, but for me the problem is that I would like to use a number of facts about product measures which come from Fubini's theorem. And of course you need to know how to integrate $d\mu$ in order to even state Fubini. So in principle it might be possible to take this approach, but you'll have to re-prove lots of stuff from scratch to avoid circularity. – Nate Eldredge Mar 2 '18 at 22:26
• My intention was to just use the Lebesgue measure on $\mathbb R^{n+1}$, why doesn't this work? – Asvin Mar 2 '18 at 22:33