Editing Integral test for convergence (section)

==Proof==
The proof uses the [[Direct comparison test|comparison test]], comparing the term <math>f(n)</math> with the integral of <math>f</math> over the intervals <math>[n-1,n)</math> and <math>[n,n+1)</math> respectively.

The monotonic function <math>f</math> is [[Continuous function|continuous]] [[almost everywhere]]. To show this, let 
:<math>D=\{ x\in [N,\infty)\mid f\text{ is discontinuous at } x\}</math>
For every <math>x\in D</math>, there exists by the [[Dense set|density]] of <math>\mathbb Q</math>, a <math>c(x)\in\mathbb Q</math> so that <math>c(x)\in\left[\lim_{y\downarrow x} f(y), \lim_{y\uparrow x} f(y)\right]</math>. 

Note that this set contains an [[Open set|open]] [[non-empty]] interval precisely if <math>f</math> is [[discontinuous]] at <math>x</math>. We can uniquely identify <math>c(x)</math> as the [[rational number]] that has the least index in an [[enumeration]] <math>\mathbb N\to\mathbb Q</math> and satisfies the above property. Since <math>f</math> is [[Monotonic function|monotone]], this defines an [[Injective function|injective]] [[Function (mathematics)|mapping]] <math>c:D\to\mathbb Q, x\mapsto c(x)</math> and thus <math>D</math> is [[Countable set|countable]]. It follows that <math>f</math> is [[Continuous function|continuous]] [[almost everywhere]]. This is [[Necessity and sufficiency|sufficient]] for [[Riemann integrability]].<ref>{{Cite journal| issn = 0002-9890| volume = 43
| issue = 7| pages = 396–398| last = Brown| first = A. B.| title = A Proof of the Lebesgue Condition for Riemann Integrability| journal = The American Mathematical Monthly| date = September 1936| jstor = 2301737| doi = 10.2307/2301737}}</ref>

Since {{math|''f''}} is a monotone decreasing function, we know that

:<math>
f(x)\le f(n)\quad\text{for all }x\in[n,\infty)
</math>

and

:<math>
f(n)\le f(x)\quad\text{for all }x\in[N,n].
</math>

Hence, for every integer {{math|''n'' ≥ ''N''}},

{{NumBlk|:|<math>
\int_n^{n+1} f(x)\,dx
\le\int_{n}^{n+1} f(n)\,dx
=f(n)</math>|{{EquationRef|2}}}}

and, for every integer {{math|''n'' ≥ ''N'' + 1}},

{{NumBlk|:|<math>
f(n)=\int_{n-1}^{n} f(n)\,dx
\le\int_{n-1}^n f(x)\,dx.
</math>|{{EquationRef|3}}}}

By summation over all {{math|''n''}} from {{math|''N''}} to some larger integer {{math|''M''}}, we get from ({{EquationNote|2}})

:<math>
\int_N^{M+1}f(x)\,dx=\sum_{n=N}^M\underbrace{\int_n^{n+1}f(x)\,dx}_{\le\,f(n)}\le\sum_{n=N}^Mf(n)
</math>

and from ({{EquationNote|3}})

:<math>
\begin{align}
\sum_{n=N}^Mf(n)&=f(N)+\sum_{n=N+1}^Mf(n)\\
&\leq f(N)+\sum_{n=N+1}^M\underbrace{\int_{n-1}^n f(x)\,dx}_{\ge\,f(n)}\\
&=f(N)+\int_N^M f(x)\,dx.
\end{align}
</math>

Combining these two estimates yields

:<math>\int_N^{M+1}f(x)\,dx\le\sum_{n=N}^Mf(n)\le f(N)+\int_N^M f(x)\,dx.</math>

Letting {{math|''M''}} tend to infinity, the bounds in ({{EquationNote|1}}) and the result follow.