1
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ It is not hard to prove that for $f : [a, b] \to [0,\infty]$, $$\oint f(x) \, dx = \int \left( \liminf_{y \to x} f(y) \right) \, dx $$ where the right-hand side is Lebesgue integral. Since $x \mapsto \liminf_{y \to x} f(y)$ is lower semicontinuous, it is Lebesgue measurable and hence the RHS makes sense. And of course the integrability of $f$ can be characterized by this function. Finally, this integral lacks linearlity: consider the sum of integrals of $\mathbf{1}_{\mathbb{Q}}$ and $\mathbb{1}_{\mathbb{R}\setminus\mathbb{Q}}$ for instance. $\endgroup$ – Sangchul Lee Feb 3 '18 at 0:28
  • $\begingroup$ @SangchulLee I don't understand your comment about $1_Q$ and $1_{R \setminus Q}$. I assume you consider these restricted to $[a,b]$. Then they are both inner-integrable. And so is their sum. $\endgroup$ – LucasSilva Feb 3 '18 at 1:43
  • 1
    $\begingroup$ Yes, all of them are integrable in the sense you describe, but $$\oint_{0}^{1}\mathbf{1}_{\mathbb{Q}}(x)\,dx+\oint_{0}^{1}\mathbf{1}_{\mathbb{R}\setminus\mathbb{Q}}(x)\,dx = 0 \neq 1 = \oint_{0}^{1}(\mathbf{1}_{\mathbb{Q}}(x) + \mathbf{1}_{\mathbb{R}\setminus\mathbb{Q}}(x))\,dx.$$ So it lacks the linearity. $\endgroup$ – Sangchul Lee Feb 3 '18 at 1:46
  • $\begingroup$ @SangchulLee Indeed. Thanks. $\endgroup$ – LucasSilva Feb 3 '18 at 1:49
  • $\begingroup$ @SangchulLee If you care to write up your comment as an answer, that would be great. In particular, it would be nice to have some explanation of how you saw so quickly that the inner integral for positive functions is the integral of $\liminf$. I'd like to understand the intuition. $\endgroup$ – LucasSilva Feb 5 '18 at 21:15
1
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.