I am trying to prove Lebesgue's criterion for Riemann integrability which states that:

A bounded function $f:[a,b]\to \mathbb R$ is Riemann integrable iff it is continuous $\lambda$-a.e. and if it is Riemann integrable then it is also Lebesgue integrable and the values of the two integrals coincide.

I started in the following way:

Take $\mathcal P_k=\{a=x_0<x_1<...<x_k=b\}$ be a sequence of partitions of $[a,b]$ such that $\mathcal P_{k+1}\supset \mathcal P_k$ and $||\mathcal P_k||\to 0$ as$k\to \infty$.

Let $L_k=\sum\limits_{i=1}^{k} m_i\chi_{(x_{i-1},x_i]}$ and $U_k=\sum\limits_{i=1}^{k} M_i\chi_{(x_{i-1},x_i]}$,where $m_i$ and $M_i$ are respectively the infimum and supremum of $f$ over $[x_{i-1},x_i]$.

Then we can write $L(f,\mathcal P_k)=\int_{[a,b]}L_kd\lambda$ and $U(f,\mathcal P_k)=\int_{[a,b]} U_kd\lambda$.Now I am stuck and need some help.Can someone show me a way?


1 Answer 1



Suppose $A=A(\epsilon)$ is a finite union of open* intervals with disjoint closures* and total measure less than $\epsilon$ that covers all points of discontinuity of $f$. Incorporate the endpoints of the intervals in $A$ into a partition. Control the error coming from $A$ by brute force using the fact that $f$ is bounded. Meanwhile $A^c$ is a finite union of compact intervals on which $f$ is continuous; partition that as fine as needed. Sum up the errors on $A$ and $A^c$ to finish.

* Here "open" is relative to the topology of $[a,b]$ itself, not that of $\mathbb{R}$.


Introduce the concept of oscillation at a point, $\omega_f(x)$. Let $D_i=\{ x : \omega_f(x) \geq a_i \}$ where $a_i$ is a sequence to be determined that decreases to zero. Then $x$ is a point of discontinuity of $f$ if and only if there exists $i$ such that $x \in D_i$. We want to show $\lambda \left ( \bigcup_{n=1}^\infty D_i \right )<\epsilon$. We'll do that by choosing our favorite $b_i>0$ with $\sum_{i=1}^\infty b_i=\epsilon$ followed by choosing $a_i$ to ensure that $\lambda(D_i)<b_i$.

Now take a partition $P_i$ with $U(f,P_i)-L(f,P_i)<c_i$ with $c_i>0$ to be determined. The subintervals of $P_i$ which have nonempty intersection with $D_i$ have some measure $d_i \geq \lambda(D_i)$. Thus it suffices to show $d_i<b_i$. These subintervals contribute at least $d_i a_i$ to $U(f,P_i)-L(f,P_i)$. Therefore $d_i a_i<c_i$ and so $d_i<\frac{c_i}{a_i}$. Thus it suffices to choose $c_i,a_i$ both going to zero such that $\frac{c_i}{a_i}<b_i$.

A way to prove that the Lebesgue integral of a (properly) Riemann integrable function is its Riemann integral starts by noting that the Riemann and Lebesgue integrals of step functions are equal (this may or may not be immediate from the definitions, it depends which definitions you chose to use). Then use the fact that finite unions of intervals have arbitrarily close measure to an arbitrary measurable set. From here you can argue that the Lebesgue integral of a simple function is arbitrarily close to the Lebesgue integral of a step function. Then you just use density of simple functions as usual.

  • $\begingroup$ What about the proof that Riemann and Lebesgue integrals coincide? $\endgroup$ Commented Apr 26, 2022 at 16:25
  • $\begingroup$ @KishalaySarkar See revised answer. $\endgroup$
    – Ian
    Commented Apr 26, 2022 at 17:53

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