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Which Functions Are Integrable, and How the Integral Behaves

Prove that continuous functions and monotone functions are integrable, then collect the everyday rules: linearity, additivity over intervals, monotonicity, and the triangle inequality for integrals.

Continuous functions are integrable

The headline result: every continuous function on a closed bounded interval is Riemann integrable. The secret ingredient is that continuity on [a,b] is automatically uniform continuity — one δ works across the whole interval, not a different δ at each point. That uniform δ is what lets us control the wiggle M_k − m_k everywhere at once.

Theorem: f continuous on [a,b]  =>  f Riemann integrable on [a,b].

Proof. Let ε > 0. We aim for a partition with U(f,P) - L(f,P) < ε.
f is continuous on the compact [a,b], hence UNIFORMLY continuous:
   there is δ > 0 such that  |x - y| < δ  =>  |f(x) - f(y)| < ε/(b - a).

Choose any partition P with mesh ‖P‖ < δ (e.g. n equal pieces, n large).
On each subinterval [x_{k-1}, x_k] f attains its max and min (Extreme Value Thm),
say at points p_k, q_k. Both lie in the piece, so |p_k - q_k| <= Δx_k < δ. Then
   M_k - m_k = f(p_k) - f(q_k) <= |f(p_k) - f(q_k)| < ε/(b - a).

Sum it up:
   U(f,P) - L(f,P) = sum (M_k - m_k) Δx_k
                   <  (ε/(b-a)) * sum Δx_k
                   =  (ε/(b-a)) * (b - a)  =  ε.
By the Riemann criterion, f is integrable.   QED
Uniform continuity tames the wiggle on every piece at once.

Monotone functions, and the bigger picture

Continuity is sufficient but not necessary. A function can jump and still be integrable. Any monotone f on [a,b] (increasing, say) is integrable even with infinitely many jumps. Using n equal pieces of width (b−a)/n, the wiggles telescope:

f increasing on [a,b], partition into n equal pieces, Δx = (b-a)/n.
For increasing f:  M_k = f(x_k),  m_k = f(x_{k-1}).  So
   U(f,P) - L(f,P) = sum_{k=1}^n ( f(x_k) - f(x_{k-1}) ) * Δx
                   = Δx * sum ( f(x_k) - f(x_{k-1}) )    (telescopes!)
                   = Δx * ( f(b) - f(a) )
                   = (b-a)(f(b) - f(a)) / n.
Given ε>0, pick n > (b-a)(f(b)-f(a))/ε. Then U - L < ε. Integrable.   QED
For a monotone function the gap telescopes to (b−a)(f(b)−f(a))/n.

How far can we push? The complete answer is the Lebesgue criterion: a bounded f on [a,b] is Riemann integrable if and only if its set of discontinuities has “measure zero” — it can be covered by intervals of arbitrarily small total length. Continuous functions (no discontinuities) and monotone functions (at most countably many jumps) both pass easily; Dirichlet's function (discontinuous everywhere) fails. We mention this as the horizon; it belongs to measure theory.

The rules you compute with

Once functions are integrable, the integral obeys the algebra you expect. These all follow from the Riemann criterion plus the sum/scaling behaviour of sup and inf.

  1. Linearity: if f, g are integrable and c is constant, then ∫(f + g) = ∫f + ∫g and ∫(c·f) = c·∫f.
  2. Additivity over intervals: for a < c < b, ∫ from a to b = ∫ from a to c + ∫ from c to b.
  3. Monotonicity: if f ≤ g pointwise on [a,b], then ∫f ≤ ∫g. In particular f ≥ 0 gives ∫f ≥ 0.
  4. Triangle inequality for integrals: |∫f| ≤ ∫|f|, the integral cousin of the triangle inequality (and |f| is integrable whenever f is).