Riemann Integrable Functions Are Lebesgue Integrable

From Darboux Sums to Lebesgue Integrability

The previous tutorial showed that Darboux sums bound the Lebesgue integral of a bounded measurable function. For a continuous function, the bounds become arbitrarily close, giving agreement of the two integrals. We now remove the continuity assumption: every bounded Riemann integrable function on a closed bounded interval is Lebesgue integrable, and the two integrals agree.

There is a point to settle before applying the Darboux Bounds Theorem: that theorem assumes Lebesgue measurability, while a Riemann integrable function is initially described using partitions and sums. We will use partitions with small upper-minus-lower sums to construct measurable step functions that approach the function outside a null set. Completeness of Lebesgue measure then gives measurability everywhere.

Throughout, let \(a<b\), and let \(f:[a,b]\to\mathbb{R}\) be bounded and Riemann integrable. For a partition \(P=\{a=x_0<x_1<\cdots<x_n=b\}\), write \(m_i\) and \(M_i\) for the infimum and supremum of \(f\) on \([x_{i-1},x_i]\). Define step functions \(l_P\) and \(u_P\) to take the values \(m_i\) and \(M_i\), respectively, on \([x_{i-1},x_i)\), and to take \(m_n\) and \(M_n\) at \(b\).

The functions \(l_P\) and \(u_P\) are measurable step functions. Their integrals are the corresponding Darboux sums, by the simple-function integral formula and the fact that the partition endpoints have measure zero. We will use partitions for which the difference on the right tends to zero.

Small Darboux Gaps Give Measurability

Proof. Since \(f\) is Riemann integrable, for every positive integer \(n\) there is a partition \(P_n\) such that \(U(f,P_n)-L(f,P_n)<2^{-2n}\). Let \(l_n=l_{P_n}\), \(u_n=u_{P_n}\), and \(h_n=u_n-l_n\). Then \(h_n\) is measurable and nonnegative, and \(\int_{[a,b]}h_n\,d\lambda<2^{-2n}\).

Consider the measurable set \(E_n=\{x\in[a,b]:h_n(x)\geq 2^{-n}\}\). On \(E_n\), the function \(h_n\) is at least \(2^{-n}\), so monotonicity of the nonnegative integral gives \(2^{-n}\lambda(E_n)\leq\int_{[a,b]}h_n\,d\lambda<2^{-2n}\). Therefore \(\lambda(E_n)<2^{-n}\), and hence \(\sum_{n=1}^{\infty}\lambda(E_n)<\infty\). By the First Borel–Cantelli Lemma, the set \(N=\limsup_{n\to\infty}E_n\) has measure zero.

If \(x\notin N\), then \(x\) belongs to only finitely many \(E_n\). Thus, for all sufficiently large \(n\), \(h_n(x)<2^{-n}\). Since \(l_n(x)\leq f(x)\leq u_n(x)\), it follows that \(0\leq f(x)-l_n(x)\leq h_n(x)<2^{-n}\) for all such \(n\). Consequently, \(l_n(x)\to f(x)\) for every \(x\notin N\).

The function \(g=\limsup_{n\to\infty}l_n\) is measurable by the measurability theorem for limsup, and it is finite because \(f\) is bounded and each \(l_n\) takes values between a bound for \(f\) and its negative. Outside \(N\), the sequence \(l_n\) converges to \(f\), so \(g=f\) there. For any open set \(O\subseteq\mathbb{R}\), the sets \(f^{-1}(O)\) and \(g^{-1}(O)\) can differ only at points of \(N\). Since \(g^{-1}(O)\) is measurable, completeness of Lebesgue measure implies that every subset of \(N\) is measurable. It follows that \(f^{-1}(O)\) is measurable. The preimage characterization of measurability proves that \(f\) is Lebesgue measurable.

The Integrals Agree

Proof. By the measurability lemma, \(f\) is Lebesgue measurable. Because \(f\) is bounded, there is a finite constant \(C\geq0\) such that \(|f(x)|\leq C\) for every \(x\in[a,b]\). The interval has finite measure \(\lambda([a,b])=b-a\), so monotonicity of the integral gives \(\int_{[a,b]}|f|\,d\lambda\leq C(b-a)<\infty\). Thus \(f\) is Lebesgue integrable.

Let \(R\) denote the Riemann integral of \(f\), and let \(I=\int_{[a,b]}f\,d\lambda\). For each positive integer \(n\), choose a partition \(P_n\) with \(U(f,P_n)-L(f,P_n)<2^{-2n}\). The Darboux Bounds Theorem applies now that \(f\) is measurable, giving \(L(f,P_n)\leq I\leq U(f,P_n)\). By the definition of the Riemann integral through lower and upper Darboux sums, \(L(f,P_n)\leq R\leq U(f,P_n)\) as well. Both \(I\) and \(R\) therefore lie in the same interval, so \(|I-R|\leq U(f,P_n)-L(f,P_n)<2^{-2n}\). This holds for every \(n\), and the bound tends to zero. Hence \(I=R\), as claimed.

Worked Examples

Why the Hypotheses Matter

The proof uses two distinct features of the setting. First, Riemann integrability supplies partitions whose Darboux gaps tend to zero. That controls the measurable step-function bounds and leads to measurability outside a null set. Second, boundedness on a finite interval ensures that \(\int|f|\,d\lambda\) is finite. Measurability alone would not provide that finiteness.

There is also a useful distinction between measurability and pointwise convergence in the proof. The lower step functions need not converge to \(f\) at every point. The small integral gaps instead show that they converge outside a null set; completeness then handles any remaining points. This is why the conclusion is Lebesgue measurability, even though the approximation initially controls the function only almost everywhere.

The theorem completes the comparison begun in the previous tutorial. For a bounded Riemann integrable function on a closed bounded interval, the Lebesgue integral is not merely available: it equals the Riemann integral. The result does not assert that every Lebesgue integrable function is Riemann integrable; the indicator of the rationals from the previous tutorial shows that the reverse implication can fail.