Phase 6 - Lesson 6.3

The Lebesgue Integral

Integrate by slicing the range, not the domain - and gain functions the Riemann integral cannot touch.

⏱ 55 min● Advanced🔗 Prereqs: 6.2

Builds on: Measures (6.2) supply the weights; here we sum function values against them.
Leads to: Expectation (Phase 7) is a Lebesgue integral against a probability measure.

Learning Objectives

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Define measurable and simple functions and integrate a simple function.
Build the Lebesgue integral of a nonnegative function via a supremum over simple functions.
Contrast Lebesgue (partition the range) with Riemann (partition the domain).
Integrate the Dirichlet function and explain why Riemann fails but Lebesgue succeeds.
Extend the integral to signed and integrable (\(L^1\)) functions.

Key Vocabulary

Measurable function: \(f\) with \(\{x:f(x)\gt a\}\in\mathcal{F}\) for every \(a\); the class we can integrate.

Simple function: A finite combination \(\sum_i c_i\mathbf{1}_{A_i}\) of indicators of measurable sets \(A_i\).

Lebesgue integral: \(\int f\,d\mu=\sup\{\int s\,d\mu: 0\le s\le f,\ s\ \text{simple}\}\) for \(f\ge0\).

Integrable function: An \(f\) with \(\int|f|\,d\mu\lt \infty\); the space is denoted \(L^1(\mu)\).

Layer-cake formula: \(\int f\,d\mu=\int_0^\infty\mu(\{f\gt t\})\,dt\) for \(f\ge0\); integration by horizontal slices.

Dirichlet function: The indicator of the rationals; Riemann-nonintegrable but Lebesgue-integrable with integral 0.

Intuition & Motivation

Intuition

Riemann integrates by chopping the x-axis into thin columns and summing heights. Lebesgue chops the y-axis: ‘how much domain has value near \(t\)?’, weighs it by the measure of that level set, and sums. Picture counting a pile of coins by sorting them into denominations (Lebesgue) rather than in the order you pick them up (Riemann). Sorting by value is robust to wild reorderings of the domain - which is exactly why the Lebesgue integral tames functions, like the indicator of the rationals, that shred a Riemann partition.

From simple functions up

Definition - Integral of a simple function

For \(s=\sum_{i=1}^n c_i\mathbf{1}_{A_i}\) with disjoint measurable \(A_i\) and \(c_i\ge0\), define \(\int s\,d\mu=\sum_{i=1}^n c_i\,\mu(A_i)\) (with the convention \(0\cdot\infty=0\)).

For a general nonnegative measurable \(f\), approximate it from below by simple functions and take the supremum:

\[\int_X f\,d\mu = \sup\Big\{\int_X s\,d\mu \;:\; 0\le s\le f,\ s\ \text{simple}\Big\}.\] (6.1)

Theorem - Every nonnegative measurable \(f\) is a limit of simple functions

There exist simple \(0\le s_1\le s_2\le\cdots\uparrow f\) pointwise (e.g. dyadic truncation of the range). Hence the supremum above is attained as an increasing limit, and the integral is well defined.

Split a signed \(f=f^+-f^-\) into positive and negative parts. If both integrals are finite (equivalently \(\int|f|\lt \infty\)), \(f\in L^1\) and \(\int f=\int f^+-\int f^-\).

Worked Example - The Dirichlet function: Riemann fails, Lebesgue wins

Let \(D(x)=\mathbf{1}_{\mathbb{Q}}(x)\) on \([0,1]\): 1 on rationals, 0 elsewhere.

Riemann: every subinterval contains both rationals and irrationals, so the lower sum is 0 and the upper sum is 1 for every partition - the Riemann integral does not exist.

Lebesgue: \(D\) is a simple function, \(\int D\,d\mu=1\cdot\mu(\mathbb{Q}\cap[0,1])+0\cdot\mu(\text{irrationals})\).

Since \(\mu(\mathbb{Q}\cap[0,1])=0\), \(\int D\,d\mu=0\). The rationals are null, so \(D=0\) almost everywhere.

Layer-cake view

For \(f\ge0\), the horizontal-slice identity \(\int f\,d\mu=\int_0^\infty\mu(\{f\gt t\})\,dt\) makes the ‘partition the range’ picture literal and is the practical way to compute many integrals and tail expectations.

Interactive: simple-function integral and the Dirichlet function

Compute a Lebesgue integral of a simple function directly from values and the measures of its level sets, then confirm the Dirichlet integral is 0.

Interactive: a function that breaks Riemann

Sample the (highly oscillatory) function below; imagine refining a domain partition and note the upper/lower sums need not agree - motivation for slicing the range instead.

Common Mistakes to Avoid

Thinking Lebesgue and Riemann give different answers when both exist - they agree; Lebesgue only extends the class.
Integrating a non-measurable function - measurability is a prerequisite.
Ignoring integrability: \(\int f\) may be \(\infty-\infty\) if both parts are infinite (then \(f\notin L^1\)).
Forgetting the convention \(0\cdot\infty=0\) when a value multiplies an infinite-measure set.

Quant Practitioner Tips

Compute nonnegative integrals via the layer-cake \(\int_0^\infty\mu(\{f\gt t\})dt\) - often the fastest route.
‘Change on a null set’ never changes an integral; exploit this freely.
To check integrability, bound \(\int|f|\) first; sign issues come later.
In probability this integral IS expectation - the same machinery, weight = probability.

Knowledge Check

Q1 Easy

The essential difference between the Lebesgue and Riemann integrals is that Lebesgue:

Partitions the range (y-axis) and weights level sets by measure

Partitions the domain into finer subintervals

Only works for continuous functions

Always gives a larger value

Q2 Medium

The Dirichlet function \(\mathbf{1}_{\mathbb{Q}}\) on \([0,1]\) has Lebesgue integral:

Undefined

Infinite

Q3 Medium

An \(f\) is in \(L^1(\mu)\) precisely when:

\(f\) is continuous

\(\int|f|\,d\mu\lt \infty\)

\(f\) is bounded

\(\int f\,d\mu=0\)

Practical Exercise

Let \(f=3\cdot\mathbf{1}_{[0,1/2)}+7\cdot\mathbf{1}_{[1/2,1]}\) on \([0,1]\) with Lebesgue measure. Compute \(\int f\,d\mu\) from the definition, then verify it agrees with the Riemann integral.

▶ Show full solution

As a simple function, \(\int f\,d\mu=3\cdot\mu([0,1/2))+7\cdot\mu([1/2,1])=3\cdot\tfrac12+7\cdot\tfrac12=5\).

The Riemann integral of this step function is the same area \(3\cdot\tfrac12+7\cdot\tfrac12=5\). The two integrals coincide whenever the Riemann integral exists; Lebesgue merely extends the class to functions like \(\mathbf{1}_{\mathbb{Q}}\) where Riemann fails.

After the reveal, answer for yourself: Which measures of the level sets did you use, and how would the answer change if you altered \(f\) on a single point?

Lesson Summary

The Lebesgue integral is built by integrating simple functions (value times measure of level set) and taking suprema/limits for general \(f\ge0\), then splitting signs for \(L^1\). It slices the range rather than the domain, agrees with Riemann when both exist, and integrates functions like the Dirichlet function that Riemann cannot - giving 0 because the rationals are null.

Retrieval Practice

Close the lesson and answer from memory before checking. This is deliberate, effortful recall - the single highest-yield study action.

▶ Show retrieval prompts & answers

Q: How is the integral of a simple function defined?
A: \(\int\sum_i c_i\mathbf{1}_{A_i}\,d\mu=\sum_i c_i\,\mu(A_i)\): each value times the measure of its level set.

Q: Why does Riemann fail on \(\mathbf{1}_{\mathbb{Q}}\) while Lebesgue gives 0?
A: Every domain subinterval mixes rationals and irrationals, so Riemann upper/lower sums are 1 and 0. Lebesgue weights by measure: \(\mu(\mathbb{Q})=0\), so the integral is 0.

Completion Checklist

I can explain the core ideas in my own words
I worked the derivations/examples by hand
I completed the interactive workbench(es)
I passed the knowledge check

Confidence / mastery rating

Personal notes

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