01 / Riemann
Partition the input interval; sample a height.Sample sum: \(S(P,f)=\sum_{i=1}^n f(\xi_i)(x_i-x_{i-1})\)
Absolute error:
Blue dots are sample points. A finite sum is an approximation, not the definition of integrability.
Lebesgue vs Riemann integration
One integral. Two ways to gather the pieces.
Riemann groups nearby inputs. Lebesgue groups equal or similar outputs, then measures where they occur. Explore how both recover the same area, and where their definitions part company.
02 / Lebesgue
Partition the values; measure each preimage.Simple-function lower sum: \(\int s\,d\mu=\sum_k a_k\,\mu(E_k)\)
Certified upper bound:
Orange columns show the simple function rounded down to band heights. The horizontal tint marks the selected value band; the orange baseline marks its preimage.
Exact integral (both definitions):
In this lab, \(0\le f\le1\) and each value band has width \(\frac1n\). The final band includes the value 1. Thus \(0\le f-s\le 1/n\) on a domain of measure 1, so \(\int s\,d\mu\le\int f\,d\mu\le\int s\,d\mu+\frac1n\). Refinement uses nested dyadic bands. Set lengths are analytic; displayed numbers are rounded.
What the pictures are actually saying
01 / Riemann
Riemann fixes a partition \(a=x_0<\cdots<x_n=b\), chooses \(\xi_i\) in each subinterval, and forms S. A bounded function is Riemann integrable when all tagged sums approach the same finite value as the largest interval width tends to zero. One successful sequence of midpoint sums is not enough.
\(S(P,f)=\sum_{i=1}^n f(\xi_i)(x_i-x_{i-1})\)
02 / Lebesgue
Lebesgue starts with a nonnegative measurable simple function \(s=\sum_k a_k\mathbf{1}_{E_k}\) on disjoint measurable sets \(E_k\). Its integral is \(\sum_k a_k\mu(E_k)\). The sets need not be intervals. For measurable \(f\ge0\), define \(\int f\,d\mu\) as the supremum of \(\int s\,d\mu\) over all simple functions with \(0\le s\le f\). The value may be infinite.
\(\int s\,d\mu=\sum_k a_k\,\mu(E_k)\)
A related picture: horizontal layers
For nonnegative measurable f, the layer-cake identity is \(\int f\,d\mu=\int_0^\infty\mu(\{x:f(x)>t\})\,dt\). This adds thickness × length of a superlevel set. It is equivalent to the integral, but is distinct from weighting disjoint value bands by their lower values. Lebesgue integration is not merely exchanging x and y.
Where Riemann stops
Let \(D(x)=1\) when x is rational, and \(D(x)=0\) when x is irrational, on \([0,1]\). This is a definition, not a curve a computer can faithfully plot.
Every interval contains both types of point.
\(L(P,D)=0\ne1=U(P,D)\)Every lower Darboux sum is 0; every upper Darboux sum is 1. Rational tags produce a sum of 1 and irrational tags a sum of 0 at every resolution. There is no Riemann integral.
The rational set has measure zero.
\(\int D\,d\mu=0\)The rationals in \([0,1]\) are countable, and each singleton has Lebesgue measure zero. Countable additivity gives \(\mu(\mathbb Q\cap[0,1])=0\). Therefore \(\int D\,d\mu=1\cdot0=0\).
Measure zero does not mean empty
Enumerate the rationals as \(q_1,q_2,\ldots\) . Given \(\varepsilon>0\), cover \(q_j\) by an interval of length \(\varepsilon/2^j\). The union covers every rational and its total length is at most \(\varepsilon\). Since \(\varepsilon\) is arbitrary, the set has measure zero. Density and measure answer different questions.
When they agree
A bounded function on a compact interval is Riemann integrable if and only if its discontinuities form a set of Lebesgue measure zero. Whenever it is Riemann integrable, it is Lebesgue integrable with the same value. Continuity everywhere is sufficient, but not necessary.
The real gain: controlling limits
Pointwise convergence alone does not let you exchange a limit and an integral. Compare a bounded sequence with a concentrating spike. All integrals here are over \([0,1]\).
Dominated convergence
If measurable \(f_n\to f\) almost everywhere and \(|f_n|\le g\) almost everywhere for a single integrable \(g\ge0\), then f is integrable and \(\int f_n\,d\mu\to\int f\,d\mu\). The same dominating function must work for every n.
Monotone convergence
If \(0\le f_n\le f_{n+1}\) are measurable and \(f_n\to f\) almost everywhere, then \(\int f_n\,d\mu\) increases to \(\int f\,d\mu\), allowing \(+\infty\). The hypotheses matter: the concentrating spikes do not form an increasing sequence.
Signed and improper integrals
For real measurable f, set \(f^+=\max(f,0)\) and \(f^-=\max(-f,0)\). A finite Lebesgue integral requires \(\int|f|\,d\mu<\infty\) and equals \(\int f^+\,d\mu-\int f^-\,d\mu\). Extended values are possible if only one part is infinite; \(\infty-\infty\) is undefined. A conditionally convergent improper Riemann integral need not be a finite Lebesgue integral.
Check your understanding
Does convergence of midpoint sums prove Riemann integrability?
No. The definition requires convergence for arbitrary tags and partitions as the mesh tends to zero. A particular sampling rule can miss the behavior that prevents integrability.
Does changing a function on a null set preserve Riemann integrability?
Not always. Start from the zero function and change its value to 1 at every rational. The Lebesgue integral stays 0, but the resulting Dirichlet function is not Riemann integrable.
Why does the spike have no integrable dominating function?
If such a function existed, dominated convergence would force its integrals to tend to 0, contradicting their constant value 1. Each spike is integrable; the missing condition concerns one common bound for the entire sequence.
Take the explanation with you
A five-page A4 companion: definitions, exact examples, the null-set argument, convergence theorems, and exercises with answers.
Download English PDF日本語 PDFSources and scope
Original exposition and computed diagrams, informed by the following university notes. This lab uses ordinary Lebesgue length on \([0,1]\); abstract measures and nonmeasurable functions are outside its scope.