Equality of Riemann and Lebesgue Integrals for Bounded Functions

From Darboux Bounds to the Lebesgue Integral

A bounded function can be Riemann integrable even when it is not continuous. The Darboux definition captures this by asking for partitions whose upper and lower sums are arbitrarily close. Each such pair of sums also comes from measurable step functions that bound the original function. The point requiring care is that small integral gaps do not imply small gaps at every point. Instead, choosing the gaps to be summable makes the bounds squeeze the function almost everywhere; completeness of Lebesgue measure then gives measurability on the whole interval.

Let $a<b$, and use the finite complete measure space $([a,b],\mathcal{L}_{[a,b]},m)$, where $\mathcal{L}_{[a,b]}$ is the restriction of the Lebesgue sigma-algebra and $m$ is Lebesgue measure. All functions considered in the theorem are bounded real-valued functions on $[a,b]$. A partition is a finite list $P=(x_0,\ldots,x_N)$ with $a=x_0<x_1<\cdots<x_N=b$. Define $m_i=\inf_{[x_{i-1},x_i]}f$ and $M_i=\sup_{[x_{i-1},x_i]}f$. The lower and upper Darboux sums are $L(f,P)=\sum_{i=1}^N m_i(x_i-x_{i-1})$ and $U(f,P)=\sum_{i=1}^N M_i(x_i-x_{i-1})$. The Darboux criterion, established in the earlier material on Darboux sums, says that $f$ is Riemann integrable exactly when, for every $\varepsilon>0$, some partition satisfies $U(f,P)-L(f,P)<\varepsilon$. Its Riemann integral is the common value of the supremum of the lower sums and the infimum of the upper sums.

Definition: Given a partition $P$, define step functions $L_P$ and $U_P$ to equal $m_i$ and $M_i$, respectively, on $[x_{i-1},x_i)$ for $i<N$, and on $[x_{N-1},b]$ for $i=N$. These are measurable finite-valued functions, and $L_P(x)\le f(x)\le U_P(x)$ for every $x\in[a,b]$. Since each cell has measure $x_i-x_{i-1}$, their Lebesgue integrals are $\int L_P\,dm=L(f,P)$ and $\int U_P\,dm=U(f,P)$. Assigning a partition point to just one neighboring cell causes no change to these integrals.

The Equality Theorem

Theorem: If $f:[a,b]\to\mathbb{R}$ is bounded and Riemann integrable, then $f$ is Lebesgue measurable on $([a,b],\mathcal{L}_{[a,b]},m)$, and $$ \int_a^b f(x)\,dx=\int_{[a,b]} f\,dm. $$ Both integrals are finite. The integral on the left is the Riemann integral, and the integral on the right is the Lebesgue integral.

Proof. For each integer $n\ge1$, the Darboux criterion gives a partition $P_n$ such that $U(f,P_n)-L(f,P_n)<2^{-n}$. Let $L_n=L_{P_n}$, $U_n=U_{P_n}$, and $h_n=U_n-L_n$. These are measurable functions, $h_n\ge0$, and $$ \int_{[a,b]}h_n\,dm=U(f,P_n)-L(f,P_n)<2^{-n}. $$ By the Monotone Convergence Theorem, applied to the nonnegative measurable partial sums $\sum_{n=1}^k h_n$, the measurable function $H=\sum_{n=1}^{\infty}h_n$ satisfies $$ \int_{[a,b]}H\,dm=\sum_{n=1}^{\infty}\int_{[a,b]}h_n\,dm\le\sum_{n=1}^{\infty}2^{-n}=1. $$ The hypotheses of that theorem hold because the partial sums increase and every $h_n$ is nonnegative and measurable. Set $N=\{x:H(x)=\infty\}$. This is measurable. For every positive integer $k$, $H\ge k\chi_N$, so monotonicity of the integral gives $k\,m(N)\le\int H\,dm\le1$. Thus $m(N)\le1/k$ for every $k$, and consequently $m(N)=0$.

For each $x\notin N$, the series $\sum_n h_n(x)$ is finite, so $h_n(x)\to0$. Because $L_n(x)\le f(x)\le U_n(x)$ and $U_n(x)-L_n(x)=h_n(x)$, it follows that $L_n(x)\to f(x)$ for every $x\notin N$. Define $g=\liminf_{n\to\infty}L_n$. The pointwise limit inferior of measurable real-valued functions is measurable, so $g$ is measurable, and $g(x)=f(x)$ on $[a,b]\setminus N$. For every open set $O\subseteq\mathbb{R}$, the symmetric difference $f^{-1}(O)\triangle g^{-1}(O)$ is contained in $N$. The measure space uses the complete Lebesgue sigma-algebra, so every subset of the null set $N$ is measurable. Therefore $f^{-1}(O)$ is measurable for every open $O$, which proves that $f$ is Lebesgue measurable.

Since $f$ is bounded and $[a,b]$ has finite measure, $f$ is Lebesgue integrable: if $|f|\le C$, then $\int |f|\,dm\le C(b-a)<\infty$. For every $n$, monotonicity of the Lebesgue integral and the pointwise bounds give $$ L(f,P_n)=\int L_n\,dm\le \int f\,dm\le \int U_n\,dm=U(f,P_n). $$ The Riemann integral also lies between these Darboux sums, by its definition as the common supremum and infimum. Thus the absolute difference between the two integrals is at most $U(f,P_n)-L(f,P_n)<2^{-n}$ for every $n$. Since $2^{-n}\to0$, that difference is zero. $\square$

The proof does not require that the chosen partitions refine one another, nor that the functions $L_n$ increase. The summable gaps give pointwise squeezing only outside a null set; completeness is the step that turns this almost-everywhere agreement into measurability of the original function everywhere.

Worked Example: The Function f(x) = x² on [0,1]

For an integer $n\ge1$, take the partition $x_k=k/n$, $k=0,\ldots,n$. Since $x^2$ is increasing on $[0,1]$, its lower and upper Darboux sums are $$ L_n=\frac1n\sum_{k=0}^{n-1}\left(\frac{k}{n}\right)^2 =\frac{(n-1)n(2n-1)}{6n^3} =\frac{(n-1)(2n-1)}{6n^2}, $$ and $$ U_n=\frac1n\sum_{k=1}^{n}\left(\frac{k}{n}\right)^2 =\frac{n(n+1)(2n+1)}{6n^3} =\frac{(n+1)(2n+1)}{6n^2}. $$ Their difference is $$ U_n-L_n =\frac{(n+1)(2n+1)-(n-1)(2n-1)}{6n^2} =\frac{6n}{6n^2} =\frac1n. $$ It tends to zero, so the function is Riemann integrable. The theorem gives measurability and equality of the two integrals. The Darboux sums converge to $1/3$: dividing the numerator in either formula by $n^2$ shows that its limit is $2/6=1/3$. Hence $\int_0^1 x^2\,dx=1/3$. Check: at $n=2$, $L_2=(1)(3)/(6\cdot4)=1/8$, $U_2=(3)(5)/(6\cdot4)=5/8$, and the gap is $1/2=1/n$; both sums approach $1/3$ as $n$ increases.

Worked Example: An Indicator with a Jump

On $[0,1]$, set $f=\chi_{[0,1/2]}$. For the partition into $n$ equal subintervals, take $n$ even. The first $n/2$ closed cells lie in $[0,1/2]$, so their infimum and supremum are both $1$. The cell immediately to the right of $1/2$ has infimum $0$ and supremum $1$, because it includes the endpoint $1/2$ when extrema are taken on the closed cell. All later cells have infimum and supremum $0$. Therefore $$ L_n=\frac n2\cdot\frac1n=\frac12,\qquad U_n=\frac n2\cdot\frac1n+\frac1n=\frac12+\frac1n. $$ The gap is $1/n$, which tends to zero. The theorem proves that $f$ is measurable and its integrals agree. Directly, $f$ is the indicator of a measurable set of measure $1/2$, so $\int f\,dm=m([0,1/2])=1/2$. Its Riemann integral is also $1/2$, as the lower and upper sums converge to that same value. Check: for $n=4$, the lower sum is $2(1/4)=1/2$, the upper sum is $2(1/4)+1/4=3/4$, and their difference is $1/4$.

Worked Example: A Signed Function on [-1,1]

Let $f(x)=x$ on $[-1,1]$, and divide the interval into $n$ equal cells of length $2/n$. Because $f$ is increasing, the lower values are the left endpoints $-1+2k/n$, $k=0,\ldots,n-1$, and the upper values are the right endpoints $-1+2k/n$, $k=1,\ldots,n$. Summing gives $$ L_n=\frac2n\sum_{k=0}^{n-1}\left(-1+\frac{2k}{n}\right) =\frac2n\left(-n+\frac{(n-1)n}{n}\right) =-\frac2n, $$ and $$ U_n=\frac2n\sum_{k=1}^{n}\left(-1+\frac{2k}{n}\right) =\frac2n\left(-n+\frac{(n+1)n}{n}\right) =\frac2n. $$ Thus $U_n-L_n=4/n\to0$, and the common limit of the sums is $0$. The theorem identifies this with the Lebesgue integral. Directly, $\int_{[-1,1]}x\,dm=0$, since the integral on $[-1,0]$ is $-1/2$ and that on $[0,1]$ is $1/2$. Check: for $n=4$, the lower sum is $-2/4=-1/2$, the upper sum is $2/4=1/2$, and their gap is $1$, which equals $4/4$.

What the Riemann Integrability Hypothesis Contributes

The conclusion is specific to bounded Riemann integrable functions on a compact interval. Boundedness and finite interval length ensure that, once measurability is established, the Lebesgue integral is finite. Riemann integrability supplies the partitions with arbitrarily small Darboux gaps. Neither ingredient should be replaced by the bare statement that a bounded function has small gaps in integral for some unrelated sequence of bounds.

A common mistake is to infer from $\int h_n\,dm\to0$ that $h_n(x)\to0$ at every point. For instance, on $[0,1]$, $h_n=\chi_{[0,1/n]}$ has integral $1/n\to0$, but $h_n(0)=1$ for every $n$. The proof above avoids that invalid inference: the gaps are chosen with $\int h_n\,dm<2^{-n}$, so their sum has finite integral and is finite almost everywhere. This yields $h_n(x)\to0$ almost everywhere, and completeness handles the exceptional null set. It would also be incorrect to apply the Monotone Convergence Theorem directly to the lower bounds $L_n$, since partitions selected independently need not make $L_n$ increase. The theorem is applied instead to the increasing partial sums of the nonnegative functions $h_n$.

Once the theorem is established, it lets us move between Darboux computations and Lebesgue integration for bounded Riemann integrable functions without changing the value. The result does not assert that every bounded function is Riemann integrable; deciding which bounded functions satisfy that condition is a separate question.

Check Your Understanding

For each question, identify the relevant bounds and check whether the hypotheses of the equality theorem apply.

For a bounded Riemann integrable $f:[a,b]\to\mathbb{R}$, why does $\int |f|\,dm\le C(b-a)$ hold when $|f|\le C$?
Why does a sequence of Darboux gaps with integrals tending to zero not, by itself, give pointwise convergence of the gaps everywhere?
For $f=\chi_{[0,1/2]}$ on $[0,1]$, compute the lower and upper sums using $n=8$ equal cells.
For $f(x)=x$ on $[-1,1]$, compute the lower and upper sums using $n=2$ equal cells.
Where does completeness of the Lebesgue measure space enter the proof of measurability?

Equality of Riemann and Lebesgue Integrals for Bounded Functions

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