A Criterion

Lingual

A standard or test by which individual things or people may be compared and judged.

Usage notes

The plural form criterions also exists, but is much less common.
The form criteria is sometimes used as a nonstandard singular form (as in a criteria, this criteria, and so on), with corresponding plural form criterias. In this use, it sometimes means “a single criterion”, sometimes “a set of criteria”.

Mathematics

Calculus

Darboux’s Criterion

In standard modern calculus, the relationship between a Riemann sum and the definite integral is not a theorem to be proven, but a **definition: the definite integral is defined as the limit of the Riemann sum as the partition becomes infinitely fine:

\[\int_a^b f(x)\,dx \equiv \lim_{\Vert{}\mathcal{P}\Vert{} \to 0} \sum_{i=1}^{n} f(x_i^*)\,\Delta x_i\]

Proving this equality, typically means one of two fundamental questions:

Existence (Integrability): proving this limit actually exists and converges to a unique finite value for a given function
Evaluation (Fundamental Theorem of Calculus): proving this limit of rectangle sums equals the anti-derivative evaluated at endpoints $F(b) - F(a)$

1. Proving the Limit Exists (Darboux’s Criterion)

To establish that the limit of Riemann sums converges uniquely, real analysis uses Darboux sums (Upper and Lower sums).

For a bounded function f on [a, b] and a partition $\mathcal{P} = {x_0, x_1, \dots, x_n}$:

Let $M_i = \sup {f(x) : x \in [x_{i-1}, x_i]}$
Let $m_i = \inf {f(x) : x \in [x_{i-1}, x_i]}$

Define the upper and lower sums:

\[U(f, \mathcal{P}) = \sum_{i=1}^n M_i \Delta x_i, \qquad L(f, \mathcal{P}) = \sum_{i=1}^n m_i \Delta x_i\]

For any choice of sample points $x_i* \in [x_{i-1}, x_i]$, the standard Riemann sum $S(f, \mathcal{P})$ is strictly bounded:

\[L(f, \mathcal{P}) \le \sum_{i=1}^n f(x_i^*) \Delta x_i \le U(f, \mathcal{P})\]

The Core Steps:

Refinement Monotonicity: Adding more points to a partition cannot increase the upper sum or decrease the lower sum. If $\mathcal{P}2$ refines $\mathcal{P}1$:

\[L(f, \mathcal{P}_1) \le L(f, \mathcal{P}_2) \le U(f, \mathcal{P}_2) \le U(f, \mathcal{P}_1)\]

Existence of Infimum and Supremum: Because upper sums are bounded below and lower sums are bounded above, we define:

Lower Integral: $\underline{\int_a^b} f(x)\,dx = \sup_{\mathcal{P}} L(f, \mathcal{P})$
Upper Integral: $\overline{\int_a^b} f(x)\,dx = \inf_{\mathcal{P}} U(f, \mathcal{P})$

Riemann Integrability Condition: A function is Riemann integrable if and only if:

\[\underline{\int_a^b} f(x)\,dx = \overline{\int_a^b} f(x)\,dx\]

Or equivalently: for every $\epsilon > 0$, there exists a partition $\mathcal{P}$ such that:

\[U(f, \mathcal{P}) - L(f, \mathcal{P}) < \epsilon\]

Case of Continuous Functions:

If $f$ is continuous on a closed, bounded interval $[a, b]$, the Heine–Cantor Theorem guarantees that $f$ is uniformly continuous. This means for any $\epsilon > 0$, we can choose $\Delta x$ small enough that $M_i - m_i < \frac{\epsilon}{b - a}$ across all intervals, giving:

\[U(f, \mathcal{P}) - L(f, \mathcal{P}) = \sum_{i=1}^n (M_i - m_i)\Delta x_i < \frac{\epsilon}{b - a} \sum_{i=1}^n \Delta x_i = \epsilon\]

By the Squeeze Theorem, every Riemann sum must converge to this unique common value.

2. Proving the Riemann Sum Equals $F(b) - F(a)$ (FTC Part 2)

If the question is why the limit of rectangle sums equals the antiderivative, that is proven via the Fundamental Theorem of Calculus (FTC) using the Mean Value Theorem (MVT).

Theorem: If $f$ is continuous on $[a, b]$ and $F'(x) = f(x)$, then:

\[\lim_{\Vert{}\mathcal{P}\Vert{} \to 0} \sum_{i=1}^n f(x_i^*) \Delta x_i = F(b) - F(a)\]

Proof:

Partition $[a, b]$ with $a = x_0 < x_1 < \dots < x_n = b$.
Express the net change $F(b) - F(a)$ as a telescoping sum:

\[F(b) - F(a) = \sum_{i=1}^n \big( F(x_i) - F(x_{i-1}) \big)\]

By the Mean Value Theorem, because $F$ is differentiable on each $[x{i-1}, x_i]$, there exists some point $c_i \in (x{i-1}, x_i)$ such that:

\[\frac{F(x_i) - F(x_{i-1})}{x_i - x_{i-1}} = F'(c_i)\]

Since $F'(x) = f(x)$ and $\Delta x_i = x_i - x_{i-1}$:

\[F(x_i) - F(x_{i-1}) = f(c_i)\Delta x_i\]

Substitute this back into the telescoping sum:

\[F(b) - F(a) = \sum_{i=1}^n f(c_i)\Delta x_i\]

Notice that the right-hand side is an exact Riemann sum where the sample points are specifically chosen to be $x_i* = c_i$.

Taking the limit as the mesh size $\Vert{}\mathcal{P}\Vert{} \to 0$:

The left side $F(b) - F(a)$ is constant, so its limit is unchanged.
Because $f$ is continuous, any valid choice of sample points converges to the same definite integral:

\[\lim_{\Vert{}\mathcal{P}\Vert{} \to 0} \sum_{i=1}^n f(x_i^*) \Delta x_i = F(b) - F(a)\]