Two Ways to Integrate on an Interval
Riemann and Lebesgue integration assign the same value to many familiar integrals, but they organize the calculation differently. Riemann integration divides the domain into small intervals and compares the function's values on each interval. Lebesgue integration, developed earlier in this course, measures the sets on which the function takes different values. Comparing these approaches clarifies both why they agree in common situations and why Lebesgue integration applies more broadly.
We work on a bounded interval \([a,b]\), where \(a<b\), with Lebesgue measure \(\lambda\). For the Riemann integral, begin with a bounded function \(f:[a,b]\to\mathbb{R}\). A partition is a finite list \(P=\{a=x_0<x_1<\cdots<x_n=b\}\). On the \(i\)-th subinterval, define the infimum and supremum of \(f\) by \(m_i=\inf_{x\in[x_{i-1},x_i]} f(x)\) and \(M_i=\sup_{x\in[x_{i-1},x_i]} f(x)\).
These are the lower and upper Darboux sums. The lower integral is the supremum of all lower sums, and the upper integral is the infimum of all upper sums. The bounded function \(f\) is Riemann integrable if these two numbers are equal; their common value is its Riemann integral. Equivalently, \(f\) is Riemann integrable if for every \(\varepsilon>0\) some partition satisfies \(U(f,P)-L(f,P)<\varepsilon\). Indeed, refining a partition cannot decrease its lower sum or increase its upper sum, and the difference between the upper and lower integrals is the infimum of the differences \(U(f,P)-L(f,P)\).
A Lebesgue integral is defined for measurable functions. Every bounded measurable function on \([a,b]\) is Lebesgue integrable: its absolute value is bounded by a constant, and the interval has finite measure. The next result makes the comparison precise, without yet assuming that the function is Riemann integrable.
Darboux Sums Bound the Lebesgue Integral
Proof. For each \(i\), let \(m_i\) and \(M_i\) be the infimum and supremum of \(f\) on the closed interval \([x_{i-1},x_i]\). Define measurable step functions \(l_P\) and \(u_P\) to equal \(m_i\) and \(M_i\), respectively, on \([x_{i-1},x_i)\); at \(b\), give them the values \(m_n\) and \(M_n\). Each point belongs to a subinterval whose closed version contains it, so \(l_P(x)\leq f(x)\leq u_P(x)\) for every \(x\in[a,b]\).
The step functions are bounded and measurable, hence integrable on this finite-measure interval. Monotonicity of the Lebesgue integral gives \(\int l_P\,d\lambda\leq\int f\,d\lambda\leq\int u_P\,d\lambda\). The intervals \([x_{i-1},x_i)\) have measures \(x_i-x_{i-1}\), and the single point \(b\) has measure zero. Using the integral of an indicator function, or equivalently the integral formula for a simple function, we obtain \(\int l_P\,d\lambda=L(f,P)\) and \(\int u_P\,d\lambda=U(f,P)\). Substituting these equalities proves the theorem.
The bounds work for signed functions as well as nonnegative ones. Their force is especially clear when a partition makes the upper and lower sums close: the Lebesgue integral must then lie in the same narrow interval as the Riemann integral.
Agreement for Continuous Functions
Proof. A continuous function on a closed bounded interval is bounded and uniformly continuous. It is also Borel measurable: the preimage of every open set is open relative to \([a,b]\). Thus \(f\) is Lebesgue integrable on \([a,b]\).
Fix \(\varepsilon>0\). Uniform continuity gives \(\delta>0\) such that \(|f(x)-f(y)|<\varepsilon/(b-a)\) whenever \(x,y\in[a,b]\) and \(|x-y|<\delta\). Choose a partition whose subinterval lengths are all less than \(\delta\). For each subinterval, the difference between its supremum and infimum is at most \(\varepsilon/(b-a)\): any two values on it differ by less than that amount, so their supremum and infimum differ by at most that amount. Therefore $$ U(f,P)-L(f,P) =\sum_{i=1}^n(M_i-m_i)(x_i-x_{i-1}) \leq \frac{\varepsilon}{b-a}\sum_{i=1}^n(x_i-x_{i-1}) =\varepsilon. $$ As \(\varepsilon\) can be made arbitrarily small, \(f\) is Riemann integrable.
Write \(R\) for its Riemann integral and \(I=\int_{[a,b]}f\,d\lambda\). For every partition \(P\), the Darboux Bounds Theorem gives \(L(f,P)\leq I\leq U(f,P)\). Since \(R\) lies between the lower and upper sums for every partition, we have \(L(f,P)\leq R\leq U(f,P)\). For every \(\varepsilon>0\), choose \(P\) so that \(U(f,P)-L(f,P)<\varepsilon\). Both \(I\) and \(R\) belong to \([L(f,P),U(f,P)]\), so \(|I-R|<\varepsilon\). This holds for every positive \(\varepsilon\), which forces \(I=R\). The theorem follows.
Worked Example: The Function \(x^2\) on \([0,1]\)
The function \(f(x)=x^2\) is continuous on \([0,1]\), so the theorem guarantees that its Riemann and Lebesgue integrals agree. To identify the shared value, the ordinary antiderivative calculation gives $$ \int_0^1 x^2\,dx=\left[\frac{x^3}{3}\right]_0^1 =\frac{1^3}{3}-\frac{0^3}{3}=\frac13. $$ Consequently, \(\int_{[0,1]}x^2\,d\lambda=1/3\) as well. The agreement theorem supplies the link between the two definitions; the antiderivative supplies the value.
A Discontinuous Example Where the Integrals Agree
Continuity is sufficient for agreement, but it is not necessary. A simple step function shows how the Darboux comparison also handles a jump. For \(c\in(0,1)\), let \(f=\mathbf{1}_{[0,c]}\) on \([0,1]\). This is measurable, and its Lebesgue integral is the measure of the set where it equals \(1\).
Worked Example: An Indicator with One Jump
By the integral-of-an-indicator theorem and the measure of an interval, $$ \int_{[0,1]}\mathbf{1}_{[0,c]}\,d\lambda =\lambda([0,c])=c. $$ For the Riemann integral, take a partition containing \(c\). On each subinterval strictly to the left of \(c\), the function is \(1\), and on each subinterval to its right it is \(0\). Only the subintervals touching the jump can have different supremum and infimum. More explicitly, for any \(\eta>0\), choose \(r,s\) with \(0<r<c<s<1\), \(c-r<\eta\), and \(s-c<\eta\), and use a partition containing \(r,c,s\). The total contribution to \(U-L\) is at most \((c-r)+(s-c)<2\eta\), since outside \([r,s]\) the function is constant on each partition interval. As \(\eta\) is arbitrary, the function is Riemann integrable.
Its Riemann integral is \(c\): the function is \(1\) on an interval of length \(c\) and \(0\) on the remainder, while changing its value at the single endpoint \(c\) does not change the integral. Thus both methods give \(c\), even though the function is not continuous at \(c\).
Lebesgue Integrability Does Not Guarantee Riemann Integrability
The reverse direction fails. Lebesgue integration can ignore changes on null sets, whereas Darboux sums detect the range of values in every subinterval, however short. The indicator of the rationals provides a decisive example. The rationals are countable and hence Lebesgue null, as established in the earlier tutorial on Null Sets.
Worked Example: The Indicator of the Rationals
Define \(g(x)=\mathbf{1}_{\mathbb{Q}\cap[0,1]}(x)\) on \([0,1]\). The function is measurable because \(\mathbb{Q}\cap[0,1]\) is measurable. Its Lebesgue integral is $$ \int_{[0,1]}g\,d\lambda =\lambda(\mathbb{Q}\cap[0,1])=0. $$ To test Riemann integrability, take any partition of \([0,1]\). Every subinterval of positive length contains both a rational and an irrational number. Therefore the supremum of \(g\) on each subinterval is \(1\), and its infimum is \(0\). It follows that $$ U(g,P)=\sum_{i=1}^n(x_i-x_{i-1})=1, \qquad L(g,P)=0 $$ for every partition \(P\). The difference \(U(g,P)-L(g,P)\) is always \(1\), so no partition makes it arbitrarily small. The function is not Riemann integrable, despite having a Lebesgue integral.
What the Comparison Establishes
The Darboux Bounds Theorem is a useful bridge: for a bounded measurable function, every partition puts its Lebesgue integral between the corresponding lower and upper sums. If those sums can be brought arbitrarily close, the two approaches must yield the same value. The continuous-function theorem demonstrates this mechanism using uniform continuity.
The rationals example shows why the converse is not automatic. Lebesgue integration can use the small measure of a set, while Riemann upper and lower sums respond to whether that set and its complement both occur in each interval. A set can be null yet dense, as the rationals are, so small measure alone does not make the Darboux sums close.
For a general bounded Riemann integrable function, the comparison theorem would give equality once measurability is known. Establishing that connection in full is the next step. The distinction to retain here is that agreement for continuous functions is proved, and the Darboux bounds apply to bounded measurable functions; neither observation by itself asserts that every Riemann integrable function is measurable.
Check Your Understanding
Use the definitions and comparison results to distinguish what each integral detects.
- How are the lower and upper Darboux sums formed from a partition?
- Why does the Lebesgue integral of a bounded measurable function lie between its lower and upper Darboux sums?
- Which property of a continuous function on a closed bounded interval lets its upper and lower sums be made arbitrarily close?
- What are the Riemann and Lebesgue integrals of the indicator of \([0,c]\) on \([0,1]\), where \(0<c<1\)?
- Why do the upper and lower Darboux sums of the indicator of the rationals on \([0,1]\) differ by \(1\) for every partition?