Riemann Integral

Assume \(f\) is a bounded function on \([a,b]\). Also assume that \(f\geq 0\) on this interval. If \(a_i\) is a finite increasing sequence such that \(a_0=a\) and \(a_n=b\), we call it a division of \([a,b]\). If \(a_i\) is a division, for \([a_i,a_{i+1}]\), we know that \(f\) is bounded there, and so it has a least upper bound. We will call that \(f^u[a_i,a_{i+1}]\). Now we can define the upper step-sum of \(f\) on a division as f^u[a_0,a_1](a_1-a_0)+...+f^u[a_{n-1},a_n](a_n-a_{n_1}). If \(f\) is bounded from above by \(M\), this sum is between than \(M[b-a]\) and \(0\).

If \(a_i\) is a division, and \(a'_j\) is a division such that for every \(i\) there is a \(j\) such that \(a'_j=a_i\), then we say that \(a'_j\) is a subdivision of \(a_i\). If \(a'_j\) is a subdivision of \(a_i\), then we can prove that the step-sum of \(f\) on \(a'\) is less or equal to than the step-sum of \(f\) on \(a\). Since both are finite, in particular there are only a finite number of \(a'_j\) not in \(a_i\). Therefore, we can prove this by induction on the number of elements in \(a'\) which are not in \(a\). If there are \(0\), the divisions are the same. Assume it is true when there are \(n\) elements in \(a'\) which are not in \(a\), and assume that \(a'\) has \(n+1\) such elements. Let \(j_0\) be an element of \(a'\) not in \(a\). We define \(a''_i=a'_i\) if \(i<j_0\) and \(a''_i=a'_{i+1}\) otherwise. Then \(a''\) only has \(n\) items, and so the step sum on \(a''\) is less than that on \(a\). However, the step sum on \(a'\) and \(a''\) only differ in that in the latter, the term \(f^u[a'_{j_0-1},a'_{j_0+1}](a'_{j_0+1}-a'_{j_0-1})\) appears while in the former, \(f^u[a'_{j_0-1},a'_{j_0}](a'_{j_0}-a'_{j_0-1})+f^u[a'_{j_0},a'_{j_0+1}](a'_{j_0+1}-a'_{j_0})\). But since \(f^u[a'_{j_0-1},a'_{j_0+1}]\geq f^u[a'_{j_0-1},a'_{j_0}],f^u[a'_{j_0},a'_{j_0+1}]\), the former is less or equal to the latter, and so the step-sum on \(a'\) is less or equal to the step-sum on \(a''\), as required.

If \(a_i\) is a division and \(a'_i\) is a different division, we can define a common subdivision of both, by induction: \(a''_0=a\), and if we have \(a''_i\), we define \(a''_{i+1}\) as the least of \(a'\) and \(a\) that is greater than \(a''_i\).

We can define the lower step-sum analogously by taking the infimum of \(f\) on the intervals. The analogous results hold, with inequality signs reversed.

We define the integral of \(f\) from \(a\) to \(b\) to be the least upper step-sum of \(f\) and the highest lower step-sum of \(f\). If they are different, \(f\) does not have an integral (or, more properly, a Riemann integral).

Any continuous function has an integral, as the difference between the upper step-sum and the lower step sum of a division is no more than \(\epsilon\) is any two elements in the division are no more than \(\delta\) apart, and \(\delta\) corresponds to \(\epsilon\) for uniform continuouity, and there are always such divisions – since if \(1/n<\delta/(b-a)\), then a_i=a+(b-a)i/n is such a division.

Assume that \(f\) is bounded. We define \(f^+=(|f|+f)/2\), \(f^-=(|f|-f)/2\). If \(f\) is bounded, so are \(f^+\) and \(f^-\), which are both non-negative. \(f=f^+-f^-\). We define the integral of \(f\) as the integral of \(f^+\) minus the integral of \(f^-\). We will note that if \(f\) is continuous, then so is \(|f|\) and then so are \(f^+\) and \(f^-\), and therefore any continuous function, not necessarily non-negative, still has an integral.

It is trivial to prove that \(\int_a^b f=\int_a^c f + \int_c^b f\) – for any division, have a subdivision that includes \(c\).

It is also trivial to prove that if \(m<f<M\) then \(m(b-a)\leq \int_a^b f\leq M(b-a)\).

Assume \(f\) is continuous. \(\int_a^{x+h} f-\int_a^x = \int_x^{x+h} f\). If \(\delta\) is such that \(0<h<\delta\) such that \(|f(x+h)-f(x)|<\epsilon\), then for such \(h\), \(0\leq f\leq h\) implies \(f(x)-\epsilon<f(x+f)<f(x)+\epsilon\), and so \(f(x)-\epsilon<(\int_x^{x+h} f)/h<f(x)+\epsilon\). Analogous proof shows the same for \(h<0\), and so the derivative of \(\int_a^x f\) is \(f(x)\) when \(f\) is continuous.

Now assume that \(f\) is a function, and \(f'\) is continuous. Then the derivative of the integral of \(f'\) and the derivative of \(f\) are the same. Therefore, the derivative of their difference is always \(0\). As a consequence of Roll’s theorem, if the derivative is identically \(0\), the difference is constant. So, the integral of the derivative of a function differs from the function by a constant. This is the fundamental theorem of calculus.