Riemann Integration Proof Workshop

Two Proof Techniques for Riemann Integrals

The equivalence of Darboux and Riemann definitions gives us flexibility: we can use whichever description makes a proof easier. In this workshop, we use Darboux gaps to prove that products of integrable functions are integrable. We then use that result to prove an inequality for integrals, by integrating a family of nonnegative squares.

Both arguments illustrate a useful proof habit. First identify what the definition requires—in the first result, a small Darboux gap. Then find a pointwise estimate that turns the desired conclusion into quantities already under control. The product estimate will use oscillations on each subinterval; the inequality proof will use nonnegativity and the fact that a quadratic polynomial cannot be negative for any real input.

A Product Oscillation Estimate

For a bounded function \(u\) on a set \(J\), its oscillation on \(J\) is \(\sup_{x,y\in J}|u(x)-u(y)|\). On a partition subinterval, this is the difference between the supremum and infimum used in the Darboux gap. To estimate the oscillation of a product, insert and subtract one mixed product:

The triangle inequality then shows why bounds on the factors matter: a small change in either factor produces a controlled change in the product. We now turn this observation into an integrability theorem.

Proof. Choose positive bounds \(B_f\) and \(B_g\) such that \(|f(x)|\leq B_f\) and \(|g(x)|\leq B_g\) for every \(x\in[a,b]\). For any subinterval \(J\subseteq[a,b]\), the displayed identity gives, for all \(x,y\in J\),

Taking the supremum over \(x,y\in J\) yields

Let \(\varepsilon>0\). By the Darboux Criterion, choose a partition \(P_g\) with \(U(g,P_g)-L(g,P_g)<\varepsilon/(2B_f)\), and a partition \(P_f\) with \(U(f,P_f)-L(f,P_f)<\varepsilon/(2B_g)\). Let \(P\) be a common refinement of these partitions. Refinement does not increase the upper sum or decrease the lower sum. Therefore,

Apply the oscillation estimate on every subinterval \(J_i\) of \(P\), multiply by its length \(\Delta x_i\), and sum. The Oscillation Formula for the Darboux Gap gives

The function \(fg\) is bounded because \(|f(x)g(x)|\leq B_fB_g\). The Darboux Criterion now implies that \(fg\) is Riemann integrable. \(\square\)

The use of a common refinement is essential to this estimate: both factors must be controlled on the same subintervals before their oscillations can be added cell by cell. The refinement results established earlier in the course let us combine the two partitions without losing either gap bound.

The Cauchy–Schwarz Inequality for Integrals

The product theorem makes it legitimate to integrate \(f^2\), \(g^2\), and \(fg\) whenever \(f\) and \(g\) are Riemann integrable. It also ensures that \((f-tg)^2\) is integrable for each real number \(t\): first \(f-tg\) is integrable by linearity, and then its product with itself is integrable. Since this square is nonnegative, order preservation of the integral gives a nonnegative number. That fact, applied for every \(t\), contains the desired inequality.

Proof. Set

All three integrals exist by the Integrability of Products theorem. Since \(f^2\geq0\) and \(g^2\geq0\), order preservation implies \(A\geq0\) and \(B\geq0\). For each \(t\in\mathbb{R}\), the function \((f-tg)^2\) is integrable and nonnegative. By linearity and order preservation,

If \(B>0\), choose \(t=C/B\). Substitution gives

Multiplication by \(B>0\) yields \(C^2\leq AB\). If \(B=0\), the inequality \(A-2tC\geq0\) holds for every real \(t\). If \(C\neq0\), choosing \(t\) with the same sign as \(C\) and sufficiently large magnitude makes \(A-2tC<0\), a contradiction. Thus \(C=0\), and \(C^2\leq AB\) follows because \(A\geq0\). In either case, \(C^2=|C|^2\), proving the theorem. \(\square\)

The zero case cannot be discarded: division by \(B\) is valid only when \(B>0\). The separate argument also shows why the proof works even if \(g\) is not identically zero as a function but its squared integral is zero. No claim about pointwise values is needed for the inequality.

A Useful Consequence and Its Application

Taking \(g(x)=1\) in the Cauchy–Schwarz inequality gives a bound on the signed integral of any Riemann integrable \(f\). The constant function is integrable, and its square has integral \(b-a\), so

This estimate compares the signed total of \(f\) with the total of its square. Positive and negative values may cancel in the signed integral, but the square remains nonnegative. The length factor \(b-a\) accounts for the constant function used in the inequality.

How to Reuse These Proofs

When proving integrability of a product, a useful sequence of steps is to find bounds for both factors, estimate the product’s oscillation on one cell, and then sum that estimate over a common refinement. This converts a new Darboux gap into a weighted sum of two gaps already known to be small.

For inequalities involving integrals, first check that every expression being integrated is Riemann integrable. Here the product theorem supplies that justification. Then look for a pointwise nonnegative expression—often a square—whose integral produces an algebraic inequality. Finally, check degenerate cases before dividing by a quantity that might be zero.

A common pitfall is to write down the integral of \((f-tg)^2\) without establishing its integrability. Pointwise nonnegativity alone does not guarantee Riemann integrability. The product theorem closes that logical gap: linearity makes \(f-tg\) integrable, and the theorem makes its square integrable. Only then can order preservation be applied.