Hi everyone, my name is Matthew, I'm an international student studying maths and physics courses in Berkeley. If anyone is in PY105, PY137B or PY112 then cool! Let me know if you want to chat about anything in those subjects :) (or in this course, of course)

Although 'Math' is literally in my name, I believe it probably has measure 0. (Though I haven't been able to prove this…)

Note: In this passage, the terms L-integral, L-measurable refer to Lebesgue integral, Lebesgue measurable etc.

Intuitively, the integral of a function $f: \mathbb{R} \rightarrow \mathbb{R}$ is the area between the graph of the function and the x-axis. In order to apply our notions of measure to integration, it makes sense that we first define integrals to output positive reals, just like a measure. Therefore we focus on functions $f: \mathbb{R} \rightarrow [0,\infty)$ which, intuitively at least, should return a positive real after integration.

It is then very natural to proceed with the definition of the Lebesgue integral, which should bring measure and integration together. We first define the Undergraph $\mathcal{U}$ of $f$: $$\mathcal{U}f = \{(x,y) \in \mathbb{R} \times [0,\infty) : 0 \leq y < f(x)\}$$

Then we simply allow the Lebesgue integral to equal: $$\int f = m (\mathcal{U}f)$$

Clearly, for the integral to be defined, $\mathcal{U}f$ must be L-measurable. In which case, $f$ is called a L-measurable function.

As Rasmus Pallisgaard has correctly pointed out, we drop the $dx$ that we usually see in Riemann integration, since we not simply adding boxes of thickness $dx$. As a side note, if $\mathcal{U}f$ is indeed measurable, $m (\mathcal{U}f) = m^* (\mathcal{U}f)$. Thus we are in a sense still just adding boxes, and likely ones that have infinitesimal dimensions.

We can introduce another property, namely that $f$ is a L-integrable function if it both L-measurable, and if its L-integral is finite. Personally, I think the name isn't very appropriate, since it is still possible to find the L-integral of a non-L-integrable function, only the answer is not finite.

A significant practical advantage L-integration has over Riemann integration is that it can work with a much wider scope of functions. For instance, Riemann integration can't handle a function that is discontinuous at an uncountable number of points, however L-integration can deal with this no problem, so long as the undergraph is L-measurable. An L-integrable function can also be unbounded, so long as the L-measure of the undergraph converges to a finite value.