Partitions and step functions
Notation (Partitions and associated step functions)
| Let \(f : [a,b] \to \mathbb{R}\) be bounded by \(M\), that is $$ | f(x) | \leq M\(on\)[a,b]$$, and let |
| be a sequence of partitions of \([a,b]\) such that \(\Gamma_{k} \subseteq \Gamma_{k+1}\) (each one refines the previous one) and whose mesh $$ | \Gamma_{k} | = \max_{i} (x_{i}^{k} - x_{i-1}^{k})\(tends to zero. Writing\)x_{i} = x_{i}^{k}\(and\)n = n_{k}$$ to lighten the notation, consider: |
-
if \(m_{i} = \inf_{[x_{i-1},x_{i}]} f(x)\), the lower step function
\[\ell_{k}(x) = \sum_{i=1}^{n-1} m_{i} \mathbf{1}_{[x_{i-1},x_{i})} + m_{n} \mathbf{1}_{[x_{n-1},x_{n}]};\] -
if \(M_{i} = \sup_{[x_{i-1},x_{i}]} f(x)\), the upper step function
\[u_{k}(x) = \sum_{i=1}^{n-1} M_{i} \mathbf{1}_{[x_{i-1},x_{i})} + M_{n} \mathbf{1}_{[x_{n-1},x_{n}]}.\]
The integrals of these simple functions are the Darboux sums:
\[\int_{[a,b]} \ell_{k}(x) \, dx = \sum_{i=1}^{n} m_{i} (x_{i} - x_{i-1}), \qquad \int_{[a,b]} u_{k}(x) \, dx = \sum_{i=1}^{n} M_{i} (x_{i} - x_{i-1}).\]Note (Monotonicity of the step functions under refinement)
Note that
\[\ell_{k}(x) \leq f(x) \leq u_{k}(x) \quad \text{for every } x \in [a,b].\]Recall that if \(\Gamma_{k} \subseteq \Gamma_{k+1}\), then
- \(\ell_{k} \leq \ell_{k+1}\);
- \(u_{k} \geq u_{k+1}\).
| Since $$ | f(x) | \leq M\(for\)x \in [a,b]\(, we have\) | \ell_{k} | \leq M\(and\) | u_{k} | \leq M\(on\)[a,b]$$. By monotonicity and boundedness, the pointwise limits |
exist; they are measurable, being limits of measurable functions, and satisfy \(\ell(x) \leq f(x) \leq u(x)\) for every \(x \in [a,b]\).
Proposition (The Darboux sums converge to the integrals of \(\ell\) and \(u\))
With the notation above,
\[\lim_{k \to \infty} \int_{[a,b]} \ell_{k}(x) \, dx = \int_{[a,b]} \ell(x) \, dx, \qquad \lim_{k \to \infty} \int_{[a,b]} u_{k}(x) \, dx = \int_{[a,b]} u(x) \, dx.\]That is, these integrals are the limit of the lower and upper Darboux sums. Moreover,
\[\int_{[a,b]} u(x) \, dx = \int_{[a,b]} \ell(x) \, dx \iff \int_{[a,b]} \big( u(x) - \ell(x) \big) \, dx = 0 \iff u = \ell \ \text{a.e.}\]Proof: The functions \(\ell_{k} + M\) and \(u_{k} + M\) are measurable, non-negative, converge pointwise to \(\ell + M\) and \(u + M\) respectively, and are dominated by the constant \(2M\), which is integrable on \([a,b]\) since \(m([a,b]) = b - a < \infty\). The dominated convergence theorem then implies that
\[\lim_{k \to \infty} \int_{[a,b]} (\ell_{k} + M) \, dx = \int_{[a,b]} (\ell + M) \, dx,\]and subtracting the constant \(M(b-a)\) from both sides (linearity) gives the claim for \(\ell_{k}\); the argument for \(u_{k}\) is identical.
For the equivalences: since \(u - \ell \geq 0\) and both are integrable (they are bounded by \(M\) on a domain of finite measure), linearity gives
\[\int_{[a,b]} u \, dx - \int_{[a,b]} \ell \, dx = \int_{[a,b]} (u - \ell) \, dx,\]so the first equivalence is immediate. The second is the corollary of Markov’s inequality: a non-negative function has zero integral if and only if it is zero a.e.
Note (If the sums agree, the Riemann integral is the Lebesgue integral)
We conclude that, if the lower and upper Darboux sums converge to the same value, then
\[u = \ell = f \quad \text{a.e.}\](since \(\ell \leq f \leq u\)), the function \(f\) is measurable, as it agrees a.e. with the measurable function \(\ell\), and therefore
\[\lim_{k \to \infty} \sum_{i=1}^{n} m_{i} (x_{i} - x_{i-1}) = \lim_{k \to \infty} \sum_{i=1}^{n} M_{i} (x_{i} - x_{i-1}) = \int_{[a,b]} f(x) \, dx,\]where the integral is the Lebesgue one. That is, when \(f\) is Riemann integrable, its Riemann integral coincides with its Lebesgue integral.
Lebesgue’s characterisation of Riemann integrability
Theorem (Lebesgue’s criterion: Riemann integrable is equivalent to continuous a.e.)
Let \(f : [a,b] \to \mathbb{R}\) be bounded. Then the following are equivalent:
- \(f\) is Riemann integrable;
- \(f\) is continuous a.e. on \([a,b]\).
In that case, the Riemann and Lebesgue integrals of \(f\) over \([a,b]\) coincide.
| Proof: We fix a sequence of partitions \(\Gamma_{k}\) with \(\Gamma_{k} \subseteq \Gamma_{k+1}\) and $$\lim_{k \to \infty} | \Gamma_{k} | = 0$$, and we use the notation of the previous subsection. |
(i) implies (ii). We begin by assuming that \(f\) is Riemann integrable; then the lower and upper sums both converge to the Riemann integral, and by the previous proposition \(u = \ell = f\) a.e. Let
\[Z = \{ \ell \neq f \} \cup \{ \ell \neq u \} \cup \{ f \neq u \} \cup \bigcup_{k=1}^{\infty} \Gamma_{k}.\]Then \(m(Z) = 0\), since it is a countable union of null sets: the first three by the above, and each \(\Gamma_{k}\) because it is finite. If \(x \notin Z\), then
\[u(x) = \ell(x) = f(x), \qquad x \notin \Gamma_{k} \ \text{for every } k \geq 1.\]We claim that \(f\) is continuous at every \(x \notin Z\). Suppose that \(f\) is not continuous at \(x\). Then there exists an \(\varepsilon > 0\) such that for every \(\delta > 0\) there exists \(x_{\delta}\) with
\[|x_{\delta} - x| < \delta \qquad \text{and} \qquad |f(x) - f(x_{\delta})| > \varepsilon.\]Given \(k\), since \(x \notin \Gamma_{k}\), there exist consecutive nodes \(x_{i-1}^{k}, x_{i}^{k}\) such that \(x \in (x_{i-1}^{k}, x_{i}^{k})\), which is open. Take \(\delta > 0\) such that
\[(x - \delta, x + \delta) \subseteq (x_{i-1}^{k}, x_{i}^{k}).\]Then \(x_{\delta} \in (x_{i-1}^{k}, x_{i}^{k})\) and, since both points belong to the same subinterval of the partition,
\[\varepsilon < |f(x) - f(x_{\delta})| \leq M_{i} - m_{i} = u_{k}(x) - \ell_{k}(x).\]Since this holds for every \(k\), letting \(k \to \infty\) we obtain
\[u(x) - \ell(x) \geq \varepsilon > 0,\]which contradicts \(u(x) = \ell(x)\). We conclude that \(f\) is continuous at every point outside the null set \(Z\), that is, \(f\) is continuous a.e.
(ii) implies (i). Assume now that
\[Z = \{ x \in [a,b] :\ f\ \text{is discontinuous at } x \}\]has measure zero. Let \(x \notin Z \cup \{a, b\}\) and \(\varepsilon > 0\). By continuity of \(f\) at \(x\), there exists \(\delta > 0\) such that
\[|x - y| < \delta \implies |f(x) - f(y)| < \frac{\varepsilon}{2}.\]Now take \(k_{0}\) such that
\[|\Gamma_{k}| < \frac{\delta}{2} \quad \text{if } k \geq k_{0}.\]| For \(k \geq k_{0}\), let \(i\) be such that \(x \in [x_{i-1}, x_{i})\). Then every point \(y\) of the closed subinterval \([x_{i-1}, x_{i}]\) satisfies $$ | y - x | \leq x_{i} - x_{i-1} \leq | \Gamma_{k} | < \delta$$, that is, |
Therefore,
\[f(x) - \frac{\varepsilon}{2} \leq f(y) \leq f(x) + \frac{\varepsilon}{2} \quad \text{for } y \in [x_{i-1}, x_{i}],\]and taking the infimum and supremum over that subinterval,
\[f(x) - \frac{\varepsilon}{2} \leq m_{i} \leq M_{i} \leq f(x) + \frac{\varepsilon}{2}.\]That is, for \(k \geq k_{0}\),
\[|u_{k}(x) - f(x)| \leq \frac{\varepsilon}{2} < \varepsilon, \qquad |\ell_{k}(x) - f(x)| \leq \frac{\varepsilon}{2} < \varepsilon.\]The above tells us that
\[\lim_{k \to \infty} \ell_{k}(x) = f(x), \qquad \lim_{k \to \infty} u_{k}(x) = f(x)\]for every \(x\) outside the null set \(Z \cup \{a,b\}\), that is, \(\ell = u = f\) a.e.; in particular \(f\) is measurable. By the dominated convergence theorem (applied as in the previous proposition, with domination by the constant \(M\) on the finite-measure domain \([a,b]\)), we have that
\[\lim_{k \to \infty} \int_{[a,b]} \ell_{k}(x) \, dx = \int_{[a,b]} f(x) \, dx, \qquad \lim_{k \to \infty} \int_{[a,b]} u_{k}(x) \, dx = \int_{[a,b]} f(x) \, dx.\]That is, the lower and upper Darboux sums both converge to the same value; in particular, given \(\varepsilon > 0\) there exists a partition whose upper sum and lower sum differ by less than \(\varepsilon\), and by Darboux’s criterion \(f\) is Riemann integrable. In conclusion, \(f\) is Riemann integrable and the Riemann and Lebesgue integrals agree.