Intuition about definition of Riemann integral

Question

A function $f: [a,b] \to \mathbb{R}$ is called Riemann integrable, with integral $K$, if the following condition is satisfied:

For every $\epsilon > 0$, there is some $\delta > 0$, such that for all partitions $P$ with $\Vert P \Vert < \delta$ and for all tags $T$ of $P$, we have $|S(f,P,T)- K| < \epsilon$.

($S(f,P,T)$ is a Riemann sum)

I don't understand intuitively why we need to include the tag requirement in the definition, and I have two questions:

(1) Can someone explain why we intuitively need this? Can't we just capture the behaviour of the function in the partitions point or in the middle of it, by letting the partitions vary? (See question (2) for an alternative definition, which I believe is weaker)

(2) Suppose I would adapt the definition:

For every $\epsilon > 0$, there is some $\delta > 0$, such that for all partitions $P$ with $\Vert P \Vert < \delta$, we have $|S(f,P)- K| < \epsilon$

where $P = (x_0, \dots x_n); S(f,P):= \sum_{i=1}^n f((x_i+x_{i-1})/2) (x_i-x_{i-1})$

Can you give me a function that is integrable in this sense but not Riemann integrable?

Answer 1

You can generalize your "weaker" definition as follows.

Let $\mathcal{I}$ denote the set of closed intervals $[a,b]$ and let $\tau : \mathcal{I} \to \mathbb{R}$ be any function such that $\tau([a,b]) \in [a,b]$.

Examples are $\tau([a,b]) = a$, $\tau([a,b]) = b$, $\tau([a,b]) = \frac{a + b}{2}$ etc.

Define $S(f,P,\tau) = \sum_{i=1}^n f(\tau([x_{i-1},x_i])) (x_i-x_{i-1})$. Let us say that $f$ is $\tau$-integrable with $\tau$-integral $K$ if the obvious condition is satisfied.

As pointed out by Paramanand Singh, the following are equivalent for a bounded function $f$:

(1) $f$ is Darboux integrable

(2) $f$ is Riemann integrable

(3) $f$ is $\tau$-integrable for all $\tau$

(4) $f$ is $\tau$-integrable for some $\tau$

$\tau$-integrability seems to be conceptually simpler than Riemann integrability because it avoids to use tags. But it should be clear that there is a substantial arbitrariness in the choice of $\tau$. It can be defined by a simple rule as in your question, but it can also be "erratic". It may even depend on $f$ if you want. For example, if $f$ is continuous, then you can take $\tau([x_{i-1},x_i]))$ to be any point of $[x_{i-1},x_i]$ at which $f \mid_{[x_{i-1},x_i]}$ attains its minimum (or maximum).

So what might be the benefit of general tags? Here are some arguments.

a) If you know that $f$ is integrable (for example if $f$ is continuous or monotonic), then you can choose suitable tags $T$ which make $S(f,P,T)$ explicitly evaluable. For example, you can prove that $\int_0^t \frac{1}{1 + x^2}dx = \arctan t$ by choosing a tag as $\xi_i \in [x_{i-1},x_i]$ such that $\frac{1}{1 + \xi_i^2} = \frac{1}{1 + x_{i-1}x_i}$. I shall not go into details and I do not claim that this is an elegant proof, but it shows that general tags can be useful.

b) If you have integrable functions $f,g$, you know that their product $fg$ is integrable. Then you may choose any two $\xi_i, \xi'_i \in [x_{i-1},x_i]$ and consider the sums

$$\Sigma_{i=1}^n f(\xi_i)g(\xi'_i)(x_i - x_{i-1}) .$$

These converge to $\int_a^b f(x)g(x)dx$ as was shown by G.A. Bliss.