Integration and series · Definition

Riemann integral

real analysisintegration

Before you start you need to know

It gives precise meaning to “the area under the graph” and shows exactly which functions admit one and which do not.

Definition

A bounded function \(f\) on \([a,b]\) is integrable if the supremum \(L(f)\) of its lower sums and the infimum \(U(f)\) of its upper sums, over partitions \(P\), coincide; that common value is its integral.

\[L(f)=\sup_P L(f,P)=\inf_P U(f,P)=U(f)=\int_a^b f(x)\,dx\]

By the property seen for Darboux sums, \(L(f)\le U(f)\) always. If they are equal, that number is the integral; if not, the function is not integrable and one speaks of the lower and upper integrals. \(f\) must be bounded because otherwise the supremum on some subinterval would be infinite and the sums would stop being real numbers.

Riemann's criterion: a bounded function is integrable if and only if for every \(\varepsilon>0\) there is a partition with \(U(f,P)-L(f,P)<\varepsilon\). It is the practical way to prove integrability, since it avoids computing suprema over all partitions. With it one sees that monotone functions and those with finitely many jumps are integrable.

Guiding example: for \(f(x)=x\) on \([0,1]\) and \(P\) with \(n\) equal parts, \(L=\tfrac12-\tfrac1{2n}\) and \(U=\tfrac12+\tfrac1{2n}\); the gap \(\tfrac1n\) tends to zero and the integral is \(\tfrac12\). The Dirichlet function is the classic counterexample: \(L=0\) and \(U=1\) for every partition, so it is not integrable although bounded.

The integral is linear, monotone and satisfies \(m(b-a)\le\int_a^b f\le M(b-a)\) if \(m\le f\le M\). A typical error is to think integrable means continuous: it does not. The next concept shows that continuity is enough, and later the fundamental theorem of calculus links this integral with the derivative.

Think about it

Why can an unbounded function not be Riemann integrable?

On some subinterval the supremum would be infinite, and the upper sum would not be a real number; there would be no finite infimum of upper sums able to equal the supremum of the lower ones.

Is a bounded function that is discontinuous at a single point integrable?

Yes: enclose the point in a subinterval of arbitrarily small length; its contribution to \(U-L\) is at most (oscillation)·(length), and the rest is continuous and contributes little.

Exercises 10

Compute the following integrals

  1. \(\displaystyle \int_0^1 x\,dx\)
  2. \(\displaystyle \int_0^2 x^2\,dx\)
  3. \(\displaystyle \int_1^4 5\,dx\)
  4. \(\displaystyle \int_0^{\pi}\sin x\,dx\)
  5. \(\displaystyle \int_{-1}^{1} x^3\,dx\)

Problems

  1. Let \(f=1\) on the rationals and \(f=0\) on the irrationals, over \([0,1]\). Is it Riemann integrable?
  2. Which of these functions on \([0,1]\) is not Riemann integrable?
  3. Let \(f(x)=0\) if \(x<\tfrac12\) and \(f(x)=1\) if \(x\ge\tfrac12\), on \([0,1]\). Compute \(\displaystyle\int_0^1 f\).
  4. A bounded function \(f\) on \([a,b]\) is integrable if and only if…
  5. If \(f\) is integrable on \([0,2]\) and \(1\le f(x)\le3\), what is the tightest interval guaranteed to contain \(\int_0^2 f\)?

References

  • Stephen Abbott, Understanding Analysis, 2nd edition, Springer, 2015. §7.2–7.3
  • Walter Rudin, Principles of Mathematical Analysis, 3rd edition, McGraw-Hill, 1976. Cap. 6, Def. 6.2 y Teorema 6.6
  • Robert G. Bartle, Donald R. Sherbert, Introduction to Real Analysis, 4th edition, Wiley, 2011. §7.2