Lebesgue-style-Lower-Riemann (or inner Riemann) integral. What is the space of integrable functions? Is it deficient?

Question

Motivation

For $f:[a,b] \to [0,\infty]$, define $$\oint f = \sup_{0 \leq s \leq f} \int s$$ where the supremum is over step functions $s$ on $[a,b]$.

Aside: If $f$ is real-valued and bounded, the above is the lower Riemann integral of $f$.

For $f:[a,b] \to [-\infty,\infty]$, define $$\oint f = \oint f^+ - \oint f^-$$ whenever one of the integrals on the right is finite. Here $f^{+} = \max(f,0)$ and $f^{-} = \max(-f,0)$.

Aside: If the supremum above is over Lebesgue measurable simple functions rather than step functions, the above is the Lebesgue integral of $f$.

Define $f:[a,b] \to [-\infty,\infty]$ to be inner-Riemann-integrable if $\oint f$ is finite. This is a kind of Lebesgue-style Lower Riemann integral. One could call it an inner Riemann integral because the graph of $f$ is being approximated from inside.

Question

What space of functions is the space of inner-Riemann-integrable functions? Of real-valued bounded inner-Riemann-integrable functions? (The later space contains the Riemann integrable functions).

Is the inner-Riemann integral deficient compared to the Riemann integral? That is, does it lack some nice properties that the Riemann integral has?

Answer 1

As per request, here is a proof of my comment with some correction of the statement.

First recall that a function $f$ is called lower-semicontinuous (l.s.c.) if $f(x) \leq \liminf_{y\to x} f(y)$ on its domain. We also mention the following lemma:

Lemma. Let $f : [a, b] \to (-\infty, \infty]$ be a function. Then for each $x \in [a, b]$,

\begin{align*} \sup\{ \phi(x) : \phi \leq f \text { and } \phi \text{ is l.s.c} \} &=\sup_{\delta:\delta > 0} \left( \inf_{y : |y-x|<\delta} f(y) \right) \\ &= \min\left\{ f(x), \liminf_{y\to x} f(y) \right\}. \end{align*}

The function defined by these common values is called the lower-semicontinuous envelope of $f$.

Then the claim is as follows:

Proposition. For each $f : [a, b] \to [0, \infty]$ we have $$ \oint_{a}^{b} f(x) \, dx = \int_{a}^{b} l(x) \, dx, $$ where $l$ is the l.s.c. envelope of $f$ and the right-hand side is Lebesgue integral.

The idea is that under integral, step functions are not so different from continuous functions. So we may replace step functions in the definition of $\oint$ by continuous functions. Then the supremum of continuous functions bounded above by $f$ is the lower-semicontinuous envelope $l$.

Proof. We first prove that

$$ \oint_{a}^{b} f(x) \, dx = \sup \left\{ \int_{a}^{b} \phi(x) \, dx : 0 \leq \phi \leq f \text{ and } \phi \in C([a,b]) \right\} \tag{*}$$

Indeed,

• Let $s_n$ be a sequence of step functions such that $\int_{a}^{b} s_n \to \oint_{a}^{b} f$. Mollifying jumps of $s_n$, we can choose $\phi_n \in C([a,b])$ such that $0\leq \phi_n \leq s_n$ and $\int_{a}^{b} \phi_n \geq \int_{a}^{b} s_n - \frac{1}{n}$. Taking limit as $n\to\infty$, it follows that $\text{[LHS of (*)]} \leq \text{[RHS of (*)]}$.

• Similarly, let $\phi_n \in C([a, b])$ be such that $0 \leq \phi_n \leq f$ and $\int_{a}^{b} \phi_n \to \text{[RHS of (*)]}$. Now there exists a lower Riemann sum of $\phi_n$ whose value is within distance of $\frac{1}{n}$ from $\int_{a}^{b}\phi_n$, and such Riemann sum can be realized as an integral of a step function $s_n$ satisfying $0 \leq s_n \leq \phi_n$. So $\int_{a}^{b} s_n \geq \int_{a}^{b} \phi_n - \frac{1}{n}$ and taking limit as $n\to\infty$ proves the opposite inequality.

Then the claim follows by proving that the l.s.c. envelope $l$ satisfies

$$l(x) = \sup\{ \phi(x) : 0 \leq \phi \leq f \text{ and } \phi \in C([a, b])\}.$$