Comprehensive Core Real Analysis Proof Examination

A Proof Examination in Riemann Integration

The previous tutorial examined how uniform limits interact with Riemann integration. This tutorial turns to a related proof question: if \(f\) is Riemann integrable and \(\varphi\) is continuous, must the transformed function \(\varphi\circ f\) also be Riemann integrable? A proof needs more than the observation that \(\varphi\) is continuous. It must connect continuity of \(\varphi\) to the Darboux gap of the composition.

We will build that connection in two stages. First, a Lipschitz bound gives a direct quantitative estimate. Then uniform continuity, together with a division of partition intervals into two classes, proves the result for every continuous transformation on the bounded range of \(f\). These arguments also provide a useful examination method: state the hypotheses, control the quantities used in the criterion, and check that each estimate applies on every relevant interval.

Throughout, let \(a<b\), and let \(f:[a,b]\to\mathbb{R}\) be bounded. We use the Darboux Criterion and the Oscillation Formula for the Darboux Gap established earlier in this course. For a partition \(P=\{x_0,\ldots,x_n\}\), write \(\omega_i(f)\) for the oscillation of \(f\) on \([x_{i-1},x_i]\). The formula says that

Thus a small Darboux gap means that the oscillations, weighted by interval lengths, have a small total.

First Proof: A Quantitative Lipschitz Estimate

Proof. For any two points \(x,y\) in a partition subinterval, the Lipschitz hypothesis gives

Taking the supremum over pairs \(x,y\) in that subinterval shows that its oscillation satisfies \(\omega_i(\varphi\circ f)\leq L\omega_i(f)\). The composition is bounded because \(f(x)\in[m,M]\) and the Lipschitz condition bounds \(\varphi\) on that interval: for any fixed \(u_0\in[m,M]\),

Multiplying the oscillation inequality on each subinterval by its length and summing, the Oscillation Formula gives

If \(L=0\), then \(\varphi\) is constant on \([m,M]\), so \(\varphi\circ f\) is a constant function and is integrable. If \(L>0\), let \(\varepsilon>0\). Since \(f\) is integrable, the Darboux Criterion gives a partition \(P\) with \(U(f,P)-L(f,P)<\varepsilon/L\). The displayed estimate makes the Darboux gap of \(\varphi\circ f\) smaller than \(\varepsilon\). Applying the Darboux Criterion proves integrability. \(\square\)

The quantitative estimate is stronger than an existence argument: it shows exactly how the partition gap changes. A small Lipschitz constant can make the transformed gap smaller; a large one may enlarge it, but by at most the stated factor.

Continuous Transformations: Separate the Difficult Intervals

A continuous function on a closed bounded interval is uniformly continuous, but it need not be Lipschitz. To handle this broader case, use uniform continuity on partition intervals where \(f\) has small oscillation. The intervals where \(f\) has large oscillation need not be controlled individually; integrability ensures their total length is small.

Proof. If \(m=M\), then \(f\) is constant, and so is \(\varphi\circ f\). Suppose \(m<M\). Since \(\varphi\) is continuous on the closed bounded interval \([m,M]\), it is bounded and uniformly continuous there. Let \(C\geq0\) satisfy \(|\varphi(t)|\leq C\) for all \(t\in[m,M]\).

If \(C=0\), then \(\varphi\circ f\) is zero and is integrable. Assume \(C>0\), and let \(\varepsilon>0\). Uniform continuity gives a number \(\delta>0\) such that, for \(u,v\in[m,M]\),

Because \(f\) is Riemann integrable, choose a partition \(P=\{x_0,\ldots,x_n\}\) such that

Call a subinterval good if its oscillation \(\omega_i(f)\) is less than \(\delta\), and bad if \(\omega_i(f)\geq\delta\). On each good interval, any two values of \(f\) differ by less than \(\delta\); uniform continuity therefore bounds the oscillation of \(\varphi\circ f\) there by \(\varepsilon/(2(b-a))\). The total contribution of all good intervals to the Darboux gap is at most

On any bad interval, the oscillation of \(\varphi\circ f\) is at most \(2C\), because its values lie between \(-C\) and \(C\). Also, since \(\omega_i(f)\geq\delta\), the total length of the bad intervals satisfies

Their contribution to the Darboux gap of \(\varphi\circ f\) is consequently less than

The Oscillation Formula now bounds the full Darboux gap by the sum of the good-interval and bad-interval contributions, which is less than \(\varepsilon\). The composition is bounded because \(|\varphi(f(x))|\leq C\). The Darboux Criterion proves that \(\varphi\circ f\) is Riemann integrable. \(\square\)

The central proof technique is the split into good and bad intervals. Uniform continuity controls the transformed oscillation where the original oscillation is small. The Darboux gap of \(f\) then ensures that the intervals where this control is unavailable occupy little total length.

How to Audit a Proof

These results suggest a practical examination routine for proofs about transformed functions:

A common error is to say only that \(\varphi\) is continuous and conclude that \(\varphi\circ f\) is integrable. Continuity alone does not explain how the Darboux gap is controlled. The proof must use uniform continuity on a bounded closed range and account for intervals on which \(f\) oscillates too much for the uniform-continuity estimate to apply. Another frequent omission is the range check: \(1/t\), for example, is continuous only on intervals that avoid zero.

The broader lesson is a useful proof habit: identify the criterion that will finish the argument, express its key quantity in terms of the hypotheses, and treat every exceptional case explicitly. Here, the oscillation formula turns integrability into a weighted estimate, while continuity of the transformation transfers that estimate to the composed function.