Integration and series · Definition

Integrability of continuous functions

real analysisintegration

Before you start you need to know

It tells us a broad class of functions can be integrated without building Darboux sums by hand, and it underlies computing integrals via antiderivatives.

Definition

Result stating that every function \(f\) continuous on a closed, bounded interval \([a,b]\) is Riemann integrable there.

\[f\text{ continuous on }[a,b]\ \Rightarrow\ \int_a^b f(x)\,dx\text{ exists}\]

The proof idea is Riemann's criterion: one partition with \(U(f,P)-L(f,P)<\varepsilon\) suffices. Uniform continuity gives a \(\delta\) that works on the whole interval; if every subinterval is shorter than \(\delta\), the oscillation of \(f\) on it is at most \(\varepsilon/(b-a)\), and summing the area differences gives a gap of at most \(\varepsilon\).

The hypotheses matter. A closed, bounded interval is what turns continuity into uniform continuity; on \((0,1]\), \(1/x\) is continuous but unbounded and not integrable. Continuity is sufficient, not necessary: a bounded function with a jump, or finitely many discontinuities, is still integrable.

Running example: \(f(x)=x^2\) on \([0,1]\) is continuous, so its integral exists immediately; computing it with sums gives \(\tfrac13\). The converse fails: the Dirichlet function is nowhere continuous and not integrable, but other discontinuous functions are integrable.

This result guarantees that \(\int_a^x f\) makes sense for every continuous \(f\) and every \(x\in[a,b]\), which is needed for the fundamental theorem of calculus, now relating that integral to the derivative.

Think about it

Why is \(f(x)=1/x\) on \((0,1]\) continuous but not Riemann integrable?

The interval is not closed and \(f\) is unbounded: suprema on subintervals near \(0\) are arbitrarily large, so there are no finite upper sums.

If \(f\) is integrable on \([a,b]\), must it be continuous?

No. A function with a jump, such as \(0\) up to \(\tfrac12\) and \(1\) afterwards, is integrable and not continuous.

Exercises 10

Compute the following integrals

  1. \(\displaystyle \int_0^1 (3x^2+1)\,dx\)

Problems

  1. Which of these functions is integrable on \([0,1]\) because it is continuous on the closed interval?
  2. The integrability theorem for continuous functions requires the domain to be:
  3. Compute \(\displaystyle\int_1^3 \frac1x\,dx\), which exists because \(1/x\) is continuous on \([1,3]\).
  4. In the proof for \(f\) continuous on \([a,b]\), which property lets us choose a single \(\delta\) for the whole interval?
  5. Let \(f\) be continuous on \([a,b]\) and \(\varepsilon>0\). If \(|x-y|<\delta\) implies \(|f(x)-f(y)|<\tfrac{\varepsilon}{b-a}\), and \(P\) has subintervals of length less than \(\delta\), what follows for \(U(f,P)-L(f,P)\)?
  6. The Dirichlet function is continuous nowhere. What can be said using the integrability theorem for continuous functions?
  7. Let \(f(x)=0\) if \(0\le x<\tfrac12\) and \(f(x)=1\) if \(\tfrac12\le x\le1\). Is it integrable on \([0,1]\), and what is its integral?
  8. Let \(f(x)=\sin(1/x)\) for \(x>0\) and \(f(0)=0\). Which statement is correct on \([0,1]\)?
  9. Let \(f\ge0\) be continuous on \([0,1]\) with \(\displaystyle\int_0^1 f=0\). What follows?

References

  • Stephen Abbott, Understanding Analysis, 2nd edition, Springer, 2015. §7.2, Teorema 7.2.9
  • Walter Rudin, Principles of Mathematical Analysis, 3rd edition, McGraw-Hill, 1976. Cap. 6, Teorema 6.8
  • Robert G. Bartle, Donald R. Sherbert, Introduction to Real Analysis, 4th edition, Wiley, 2011. §7.2