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We define the Lebesgue integral of non-negative functions as:

$$\int f(x)dx=\sup_g\int g(x)dx$$

where the sup is taken over all measurable functions $g$ such that $0\leq g\leq f$ and $g$ bounded, $m(supp(g))<\infty$.

My Question:

  1. In this definition, is it inherent that the Lebesgue integral is approximating the volume of a function from "below" (the condition that $0\leq g \leq f)$?

  2. Is this why in a sense the Lebesgue integral covers a larger class of functions than the Riemann since for the Riemann, we need both the upper sum and lower sum to coincide, which loosely speaking, is tantamount to approximating the volume from both above and below?

Frank Swanton
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    For bounded functions on sets of finite measure the Lebesgue integral can be computed both in terms of an "upper" and "lower" integral as in the Riemann case. For unbound functions and/or unbounded sets the definition in terms of the supremum over lower approximations is used. Of course this extends the Lebesgue integral beyond cases where the Riemann integral can be defined. See the discussion here' – RRL Oct 08 '19 at 18:43

1 Answers1

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Actually there is a standard theorem in real analysis saying that, for a bounded domain $D$ and bounded function $f$, \begin{align*} \sup_{\varphi\leq f}\int_{D}\varphi=\inf_{\psi\geq f}\int_{D}\psi, \end{align*} the running variables $\varphi,\psi$ are all simple functions. You may figure out this theorem in, for example, Royden's book. In view of this theorem, it still gives you some sort of interpretation that one can also approximate the integral of bounded $f$ under bounded domain from above.

I think the crucial difference is that one really does the partition of the $y$-axis rather than the $x$-axis which the latter is the essence of Riemann integrability. Now assume that $f[0,1]\subseteq[0,1]$ and we are dealing with the integral on the bounded domain $[0,1]$, and the range of $f$ lies in $[0,1]$. Partitioning through $y$-axis gives you that,
\begin{align*} \sum_{i=1}^{n}\dfrac{i}{n}\chi_{f^{-1}[(i-1)/n,i/n]}\geq f, \end{align*} those $f^{-1}[(i-1)/n,i/n]$ are in some sense of the grids on the $y$-axis because one does the partition of the range of $f$, in this case, which is $[0,1]$ into $[(i-1)/n,i/n]$ and pulls back to the $x$-axis under the operation $f^{-1}$.

user284331
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