“Ito-Riemann” integration

Question

Is there a real-valued function on a closed bounded nonempty interval that is not Riemann integrable, but is "Ito-Riemann" integrable, that is: if the sampling point is always the left-end point of the subinterval in the partition?

Answer 1

Here's a simple example. Define $f\colon[0,1]\to\mathbb{R}$ by $f(x)=1/\sqrt{1-x}$ for $x < 1$ and $f(1)=0$. This is unbounded, so is not Riemann integrable. To see that $f$ is integrable when you restrict to left-hand Riemann sums, it is enough that $f$ is nonnegative and Riemann integrable on each sub-interval $[0,a]$ for $0 < a < 1$ (which is the case as it is continuous on $[0,1)$), and that the indefinite integral $I\equiv\int_0^1f(x)dx=\lim_{a\uparrow1}\int_0^af(x)dx$ is finite.

Let $a\in(0,1)$ be such that $I_a\equiv\int_0^af(x)dx\ge I-\epsilon/2$. Then, by Riemann integrability on $[0,a]$, choose $\delta > 0$ so that the Riemann sum of $f$ on any partition of $[0,a]$ of mesh less than $\delta$ is within $\epsilon/2$ of $I_a$. Letting $P=\{0=x_0 < x_1 < \cdots < x_n=1\}$ be a partition of $[0,1]$ of mesh less than $\delta$, set $y_k=\min(x_k,a)$. Then, by non-negativity of $f$, $f(x_{k-1})(x_k-x_{k-1})\ge f(y_{k-1})(y_k-y_{k-1})$ $$ I+\epsilon/2\ge I_a+\epsilon/2\ge\sum_{k=1}^nf(x_{k-1})\left(x_k-x_{k-1}\right)\ge\sum_{k=1}^nf(y_{k-1})(y_k-y_{k-1})\ge I_a-\epsilon/2\ge I-\epsilon. $$ Taking $\epsilon$ arbitrarily small, the left-hand Riemann sums converge to $I$.

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For bounded functions, on the other hand, it can be shown that Riemann integrability is equivalent to "left-hand" Riemann integrability, which I'll show now.

In fact there's two definitions of the left-hand Riemann integral which could be used. If $P=\{0=x_0 < x_1 <\cdots < x_n\}$ is a partition of the closed interval $[a,b]$ I'll write $I_P(f)$ for the left-hand Riemann sum $\sum_{k=1}^nf(x_{k-1})(x_k-x_{k-1})$. Then for a fixed real number $I$ there are two alternative definitions for the left-hand Riemann integral to exist and equal $I$ (all partitions are taken to be partitions of $[a,b]$).

Definition 1: For each $\epsilon > 0$ there exists a $\delta > 0$ such that $\lvert I_P(f)-I\rvert < \epsilon$ for every partition $P$ of mesh less than $\delta$.

Definition 2: For each $\epsilon > 0$ there exists a partition $P$ such that every partition $Q$ refining $P$ satisfies $\lvert I_Q(f)-I\rvert < \epsilon$.

First, the implications "Riemann integrable ⇒ integrable acc. to def 1 ⇒ integrable acc. to def 2" follow from the definitions. It just needs to be shown here that the implication "integrable acc. to def 2 ⇒ Riemann integrable" holds for bounded functions. So, supposing $f$ is bounded, there exists an $L > 0$ with $\lvert f(x)-f(y)\rvert\le L$ for all $x,y\in[a,b]$.

For each fixed $c > 0$ let $S_c$ be the set of points $x\in[a,b]$ with $\limsup_{y\to x}\lvert f(y)-f(x)\rvert > c$. Using Lebesgue's condition for Riemann integrability, it just needs to be shown that $S_c$ has zero measure.

For any $\epsilon > 0$ let $P$ be a partition with $\lvert I_Q(f)-I\rvert < \epsilon$ for any partition $Q$ refining $P$. Also, choose any $\alpha\in(0,1)$. If $P=\{a=x_0 < x_1 < \cdots < x_n=b\}$ then, for each $k=1,\ldots,n$, let $\tilde x_k=\alpha x_{k-1}+(1-\alpha)x_k$ and $s_k=\sup\{\lvert f(x)-f(x_{k-1})\rvert\colon x\in[x_{k-1},\tilde x_k]\}$. Note that $S_c$ is disjoint from $(x_{k-1},\tilde x_k)$ whenever $s_k \le 2c$. So, the outer Lebesgue measure of $S_c$ is bounded by $$ \lvert S_c\rvert\le\sum_{k=1}^n1_{\{s_k > 2c\}}(\tilde x_k- x_{k-1})+\sum_{k=1}^nL(x_k-\tilde x_{k-1}) \le\sum_{k=1}^n\frac{s_k}{2c}(x_k-x_{k-1})+\alpha L(b-a). $$ Now we need to show that $s_k$ are small in an average sense. We can refine the partition $P$ by introducing a new point $y_k$ in each of the intervals $[x_{k-1},\tilde x_k]$, to get a new partition $Q$. Choosing $y_k=x_{k-1}$ whenever necessary, we assume that $f(y_k)\ge f(x_{k-1})$. Then, $$ 2\epsilon > I_Q(f)-I_P(f)=\sum_{k=1}^n(f(y_k)-f(x_{k-1})(x_k-y_k)\ge\alpha\sum_{k=1}^n(f(y_k)-f(x_{k-1}))(x_k-x_{k-1}). $$ Taking the supremum over the set of possible choices of $y_k$ gives $\sum_k\sup_{y\in[x_{k-1},x_k]}(f(y)-f(x_{k-1}))(x_k-x_{k-1})$ bounded by $2\epsilon/\alpha$. Applying the same result to $-f$ gives $$ \sum_{k=1}^ms_k(x_k-x_{k-1})\le4\epsilon/\alpha. $$

Plugging this into the inequality above, $$ \lvert S_c\rvert\le\frac{2\epsilon}{\alpha c}+\alpha L(b-a). $$ Making $\alpha$ small, the second term on the right hand side can be made as small as we like then, by taking $\epsilon$ small enough, the first term on the right hand side can be made arbitrarily small. So, $S_c$ has zero outer Lebesgue measure as required.