The Central Idea
The construction proceeds in stages:
\[ \boxed{ \text{indicator functions} \longrightarrow \text{simple functions} \longrightarrow \text{nonnegative measurable functions} \longrightarrow \text{signed integrable functions} }. \]Each stage is built from the previous one, preserving monotonicity and compatibility with measure.
Why Lebesgue Integration?
Riemann integration works extremely well for continuous functions and many familiar discontinuous functions. Modern analysis, however, needs stronger control over limits and exceptional sets.
Lebesgue integration is designed to:
- integrate many highly discontinuous functions;
- ignore changes on sets of measure zero;
- support strong convergence theorems;
- organize functions into spaces such as \(L^1\) and \(L^2\);
- connect naturally with probability and functional analysis.
The Measure-Space Setting
Work on a measure space
\[ (X,\mathcal{M},\mu). \]Here \(X\) is the underlying set, \(\mathcal{M}\) is a sigma-algebra of measurable subsets, and \(\mu\) is a measure.
Functions to be integrated must be measurable with respect to this structure.
Indicator Functions
For a measurable set \(E\subseteq X\), define
\[ \mathbf{1}_E(x) = \begin{cases} 1, & x\in E,\\ 0, & x\notin E. \end{cases} \]The integral is defined by
\[ \boxed{ \int_X \mathbf{1}_E\,d\mu = \mu(E) }. \]Measure therefore becomes a special case of integration.
Scaled Indicator Functions
If \(a\ge0\), then
\[ \int_X a\mathbf{1}_E\,d\mu = a\mu(E). \]The constant \(a\) acts as a height, while \(\mu(E)\) measures the size of the set on which that height occurs.
Simple Functions
A nonnegative simple function takes only finitely many values and can be written as
\[ \phi = \sum_{k=1}^{n}a_k\mathbf{1}_{E_k}, \]where \(a_k\ge0\) and the \(E_k\) are measurable.
We may choose the \(E_k\) to be pairwise disjoint.
Integral of a Simple Function
If
\[ \phi = \sum_{k=1}^{n}a_k\mathbf{1}_{E_k} \]with pairwise disjoint measurable sets \(E_k\), define
\[ \boxed{ \int_X \phi\,d\mu = \sum_{k=1}^{n}a_k\mu(E_k) }. \]Example: A Simple Function
On \(\mathbb{R}\) with Lebesgue measure, let
\[ \phi = 2\mathbf{1}_{[0,1]} + 5\mathbf{1}_{(1,3]}. \]Then
\[ \int_{\mathbb{R}}\phi\,dm = 2(1)+5(2) = \boxed{12}. \]Approximating Nonnegative Measurable Functions
Every measurable function
\[ f:X\to[0,\infty] \]can be approximated from below by an increasing sequence of nonnegative simple functions:
\[ 0\le\phi_1\le\phi_2\le\cdots\le f \]with
\[ \phi_n(x)\to f(x). \]Integral of a Nonnegative Measurable Function
Define
\[ \boxed{ \int_X f\,d\mu = \sup \left\{ \int_X\phi\,d\mu: 0\le\phi\le f,\; \phi\text{ simple} \right\} }. \]The integral is therefore the largest integral obtainable from simple functions lying below \(f\).
The Value \(+\infty\) Is Allowed
For a nonnegative measurable function,
\[ \int_X f\,d\mu \]may equal \(+\infty\).
Thus being measurable does not automatically mean being integrable in the finite sense.
Positive and Negative Parts
For a real-valued measurable function \(f\), define
\[ f^+=\max(f,0), \qquad f^-=\max(-f,0). \]Then
\[ \boxed{f=f^+-f^-} \]and
\[ \boxed{|f|=f^++f^-}. \]Integral of a Signed Function
When the expression is well defined, set
\[ \boxed{ \int_X f\,d\mu = \int_X f^+\,d\mu - \int_X f^-\,d\mu }. \]The expression is not defined when both terms are infinite, because
\[ \infty-\infty \]is indeterminate.
Lebesgue Integrability
A measurable function \(f\) is Lebesgue integrable if
\[ \boxed{ \int_X |f|\,d\mu<\infty }. \]Equivalently, both
\[ \int_X f^+\,d\mu \]and
\[ \int_X f^-\,d\mu \]are finite.
The Space \(L^1\)
The integrable functions form the space
\[ L^1(X,\mu). \]Functions equal almost everywhere are identified with one another.
The \(L^1\) norm is
\[ \boxed{ \|f\|_1 = \int_X |f|\,d\mu }. \]Equality Almost Everywhere
If
\[ f=g \quad\text{a.e.}, \]then, whenever the integrals are defined,
\[ \int_X f\,d\mu = \int_X g\,d\mu. \]In particular,
\[ \|f-g\|_1=0. \]This explains why elements of \(L^1\) are equivalence classes modulo equality almost everywhere.
Linearity
If \(f,g\in L^1(X,\mu)\) and \(a,b\in\mathbb{R}\), then
\[ \boxed{ \int_X(af+bg)\,d\mu = a\int_Xf\,d\mu + b\int_Xg\,d\mu }. \]Monotonicity
If
\[ f\le g \quad\text{a.e.} \]and the integrals are properly defined, then
\[ \boxed{ \int_X f\,d\mu \le \int_X g\,d\mu }. \]The Integral Triangle Inequality
If \(f\in L^1\), then
\[ \boxed{ \left| \int_X f\,d\mu \right| \le \int_X|f|\,d\mu }. \]Integration over a Measurable Set
For measurable \(E\subseteq X\), define
\[ \boxed{ \int_E f\,d\mu = \int_X f\mathbf{1}_E\,d\mu }. \]In particular,
\[ \int_E1\,d\mu = \mu(E). \]Changing a Function on a Null Set
If two measurable functions differ only on a null set, then they are equal almost everywhere and have the same Lebesgue integral whenever the integral is defined.
This makes Lebesgue integration stable under negligible pointwise changes.
The Monotone Convergence Theorem
Suppose \(f_n\) are nonnegative measurable functions satisfying
\[ f_1\le f_2\le\cdots \]and
\[ f_n\to f \quad\text{a.e.} \]Then
\[ \boxed{ \int_X f\,d\mu = \lim_{n\to\infty} \int_Xf_n\,d\mu }. \]The value \(+\infty\) is permitted.
Why Monotone Convergence Matters
The theorem permits limit and integral to be interchanged without requiring a separate dominating function.
Its strength comes from the monotone structure:
- each \(f_n\) is measurable;
- each \(f_n\) is nonnegative;
- the sequence increases to its limit.
Truncation and Monotone Convergence
For nonnegative measurable \(f\), define
\[ f_n=\min(f,n). \]Then
\[ f_n\uparrow f. \]Hence
\[ \boxed{ \int_X f\,d\mu = \lim_{n\to\infty} \int_X\min(f,n)\,d\mu }. \]Fatou's Lemma
If \(f_n\ge0\) are measurable, then
\[ \boxed{ \int_X \liminf_{n\to\infty}f_n\,d\mu \le \liminf_{n\to\infty} \int_Xf_n\,d\mu }. \]Fatou's Lemma provides one-sided control even when the sequence is not monotone.
The Dominated Convergence Theorem
Suppose \(f_n\to f\) almost everywhere and each \(f_n\) is measurable. Suppose also that there exists \(g\in L^1(X,\mu)\) such that
\[ |f_n|\le g \]almost everywhere for every \(n\).
Then \(f\in L^1\) and
\[ \boxed{ \lim_{n\to\infty} \int_Xf_n\,d\mu = \int_Xf\,d\mu }. \]The Dominating Function Must Be Uniform Across the Sequence
The theorem requires one integrable function \(g\) satisfying
\[ |f_n|\le g \]for every \(n\).
Choosing a different bound \(g_n\) for each \(f_n\) does not by itself meet the hypothesis.
Dominated Convergence Gives \(L^1\) Convergence
Under the hypotheses of Dominated Convergence,
\[ |f_n-f| \le |f_n|+|f| \le 2g \]almost everywhere.
Hence
\[ \boxed{ \int_X|f_n-f|\,d\mu\to0 }. \]Equivalently,
\[ \|f_n-f\|_1\to0. \]Example: Dominated Convergence
On \([0,1]\), let
\[ f_n(x)=x^n. \]For \(0\le x<1\),
\[ x^n\to0, \]while \(f_n(1)=1\). Thus
\[ f_n\to0 \quad\text{a.e.} \]Also,
\[ |f_n|\le1, \]and \(1\in L^1([0,1])\). Therefore,
\[ \boxed{ \lim_{n\to\infty} \int_0^1x^n\,dx = 0 }. \]Bounded Convergence on Finite Measure Spaces
If \(\mu(X)<\infty\), \(f_n\to f\) almost everywhere, and
\[ |f_n|\le M \]for one constant \(M\), then the constant function \(M\) belongs to \(L^1\). Dominated Convergence therefore applies.
Absolute Continuity of the Integral
If \(f\in L^1(X,\mu)\), then for every \(\varepsilon>0\), there exists \(\delta>0\) such that
\[ \mu(E)<\delta \]implies
\[ \boxed{ \int_E|f|\,d\mu<\varepsilon }. \]Small-measure sets therefore carry arbitrarily little \(L^1\) mass.
Markov's Inequality
If \(f\ge0\) is integrable and \(t>0\), then
\[ t\mathbf{1}_{\{f>t\}} \le f. \]Integrating gives
\[ \boxed{ \mu(\{f>t\}) \le \frac1t \int_Xf\,d\mu }. \]This inequality connects the size of a function with the measure of its large-value sets.
Riemann Integrable Functions and Lebesgue Integration
Every bounded Riemann integrable function on a closed interval is Lebesgue integrable there, and the two integrals agree:
\[ \boxed{ \int_a^bf(x)\,dx_{\mathrm{Riemann}} = \int_{[a,b]}f\,dm }. \]Lebesgue Integration Reaches Further
The converse is false.
Consider
\[ f=\mathbf{1}_{\mathbb{Q}\cap[0,1]}. \]This function is discontinuous everywhere, so it is not Riemann integrable. But \(\mathbb{Q}\cap[0,1]\) has measure zero, so
\[ \boxed{ \int_{[0,1]}f\,dm=0 }. \]The Lebesgue Criterion for Riemann Integrability
A bounded function on \([a,b]\) is Riemann integrable if and only if its set of discontinuities has Lebesgue measure zero.
Measure theory therefore gives an exact characterization of bounded Riemann integrability.
Integration of Nonnegative Series
If \(f_n\ge0\) are measurable, then the partial sums increase:
\[ \sum_{k=1}^{N}f_k \uparrow \sum_{k=1}^{\infty}f_k. \]Monotone Convergence gives
\[ \boxed{ \int_X \sum_{n=1}^{\infty}f_n\,d\mu = \sum_{n=1}^{\infty} \int_Xf_n\,d\mu }. \]Absolute Summability and Termwise Integration
If
\[ \sum_{n=1}^{\infty} \int_X|f_n|\,d\mu < \infty, \]then the series \(\sum f_n\) converges absolutely almost everywhere, defines an \(L^1\) function, and may be integrated term by term:
\[ \boxed{ \int_X \sum_{n=1}^{\infty}f_n\,d\mu = \sum_{n=1}^{\infty} \int_Xf_n\,d\mu }. \]Convergence in \(L^1\)
We say
\[ f_n\to f \quad\text{in }L^1 \]if
\[ \boxed{ \|f_n-f\|_1 = \int_X|f_n-f|\,d\mu \to0 }. \]This measures total integrated error rather than pointwise error.
\(L^1\) Convergence Controls Integrals
If \(f_n\to f\) in \(L^1\), then
\[ \left| \int_Xf_n\,d\mu - \int_Xf\,d\mu \right| \le \int_X|f_n-f|\,d\mu. \]Therefore,
\[ \boxed{ \int_Xf_n\,d\mu \to \int_Xf\,d\mu }. \]Almost-Everywhere Convergence Does Not Imply \(L^1\) Convergence
On \([0,1]\), define
\[ f_n(x) = n\mathbf{1}_{(0,1/n)}(x). \]Then
\[ f_n\to0 \quad\text{a.e.} \]but
\[ \int_0^1|f_n|\,dx = n\left(\frac1n\right) = 1. \]Thus
\[ \boxed{ f_n\not\to0 \text{ in }L^1 }. \]\(L^1\) Convergence Implies Convergence in Measure
For every \(\varepsilon>0\),
\[ \mu(\{|f_n-f|>\varepsilon\}) \le \frac1\varepsilon \|f_n-f\|_1. \]Hence
\[ f_n\to f \text{ in }L^1 \]implies
\[ \boxed{ f_n\to f \text{ in measure} }. \]Completeness of \(L^1\)
The space \(L^1(X,\mu)\) is complete under
\[ \|f\|_1=\int_X|f|\,d\mu. \]Every \(L^1\)-Cauchy sequence therefore converges in \(L^1\) to an element of \(L^1\).
In functional-analysis language,
\[ \boxed{L^1\text{ is a Banach space}}. \]Complex-Valued Functions
If \(f=u+iv\), define
\[ \int_Xf\,d\mu = \int_Xu\,d\mu + i\int_Xv\,d\mu \]whenever the real and imaginary parts are integrable.
Integrability is equivalently controlled by
\[ \int_X|f|\,d\mu<\infty. \]Common Error: Measurable Does Not Mean Integrable
A measurable function may have infinite integral. To conclude
\[ f\in L^1, \]verify
\[ \int_X|f|\,d\mu<\infty. \]Common Error: Ignoring \(\infty-\infty\)
The signed integral
\[ \int f = \int f^+ - \int f^- \]is not defined when both terms equal \(+\infty\).
Common Error: Applying Dominated Convergence without Domination
Almost-everywhere convergence alone does not justify
\[ \lim\int f_n = \int\lim f_n. \]Dominated Convergence requires one integrable \(g\) satisfying
\[ |f_n|\le g \]almost everywhere for all \(n\).
Common Error: Using a Nonintegrable Dominator
A bound
\[ |f_n|\le g \]is not enough if
\[ \int_Xg\,d\mu=\infty. \]The dominating function must belong to \(L^1\).
Common Error: Confusing Convergence Modes
Pointwise convergence, almost-everywhere convergence, convergence in measure, and convergence in \(L^1\) are different concepts.
Each has its own definition and its own transfer theorems. One should never replace one with another without checking the required hypotheses.
A Reliable Procedure for Integrating a Simple Function
- Write the function as constants times indicator functions.
- Make the measurable pieces pairwise disjoint if necessary.
- Find the measure of each piece.
- Multiply each function value by the measure of its piece.
- Add the contributions.
A Reliable Procedure for Testing \(L^1\) Integrability
- Verify measurability.
- Form \(|f|\).
- Compute or estimate \(\int|f|\).
- If the result is finite, conclude \(f\in L^1\).
- If it is infinite, the function is not \(L^1\)-integrable.
A Reliable Procedure for Using Monotone Convergence
- Verify measurability.
- Verify nonnegativity.
- Check \(f_n\le f_{n+1}\) almost everywhere.
- Identify the a.e. limit \(f\).
- Apply \[ \int f=\lim\int f_n. \]
A Reliable Procedure for Using Dominated Convergence
- Identify the candidate limit \(f\).
- Verify \(f_n\to f\) almost everywhere.
- Find one function \(g\) with \(|f_n|\le g\) for every \(n\).
- Verify \(g\in L^1\).
- Conclude \(f\in L^1\).
- Interchange limit and integral.
Example: Computing an \(L^1\) Norm
Let \(f(x)=x\) on \([-1,2]\). Then
\[ \|f\|_1 = \int_{-1}^{2}|x|\,dx. \]Split at \(0\):
\[ \int_{-1}^{0}(-x)\,dx + \int_0^2x\,dx = \frac12+2. \]Therefore,
\[ \boxed{ \|f\|_1=\frac52 }. \]Practice 1: Indicator Integral
Compute
\[ \int_{\mathbb{R}} \mathbf{1}_{[2,5]}\,dm. \]Practice 2: Simple Function
Let
\[ \phi = 3\mathbf{1}_{[0,2]} + 4\mathbf{1}_{(2,5]}. \]Compute \(\int\phi\,dm\).
Practice 3: Positive and Negative Parts
For \(f(x)=x\) on \([-2,3]\), describe \(f^+\) and \(f^-\).
Practice 4: \(L^1\) Integrability
Determine whether
\[ f(x)=e^{-x} \]belongs to \(L^1([0,\infty))\).
Practice 5: Infinite Integral
Determine whether the constant function \(f(x)=1\) belongs to \(L^1(\mathbb{R})\).
Practice 6: Monotone Convergence
On \([0,\infty)\), let
\[ f_n=\mathbf{1}_{[0,n]}. \]Determine the pointwise limit and the limit of the integrals.
Practice 7: Dominated Convergence
On \([0,1]\), let
\[ f_n(x)=\frac{x}{n}. \]Use Dominated Convergence to evaluate
\[ \lim_{n\to\infty}\int_0^1f_n(x)\,dx. \]Practice 8: \(L^1\) Convergence
On \([0,1]\), let \(f_n(x)=1/n\). Determine whether \(f_n\to0\) in \(L^1\).
Practice 9: A.E. but Not \(L^1\)
On \([0,1]\), let
\[ f_n(x)=n\mathbf{1}_{(0,1/n)}(x). \]Verify that \(f_n\to0\) almost everywhere but not in \(L^1\).
Practice 10: Identify the Error
A student claims:
Since \(f_n\to f\) almost everywhere, we may always move the limit through the integral.
Explain why the claim is false.
Practice Answers
Practice 1
The integral equals the measure of the interval:
\[ \boxed{5-2=3}. \]Practice 2
The two pieces have lengths \(2\) and \(3\):
\[ 3(2)+4(3)=18. \]Therefore,
\[ \boxed{18}. \]Practice 3
We have
\[ f^+(x) = \begin{cases} 0, & x<0,\\ x, & x\ge0, \end{cases} \]and
\[ f^-(x) = \begin{cases} -x, & x<0,\\ 0, & x\ge0. \end{cases} \]Practice 4
Since
\[ \int_0^\infty e^{-x}\,dx=1, \]we have
\[ \boxed{ e^{-x}\in L^1([0,\infty)) }. \]Practice 5
We have
\[ \int_{\mathbb{R}}1\,dm = m(\mathbb{R}) = \infty. \]Therefore,
\[ \boxed{ 1\notin L^1(\mathbb{R}) }. \]Practice 6
The functions increase pointwise to \(1\) on \([0,\infty)\). Also,
\[ \int_0^\infty f_n\,dm=n. \]Therefore,
\[ \boxed{ \lim_{n\to\infty}\int f_n\,dm = \infty = \int_0^\infty1\,dm }. \]Practice 7
We have \(f_n\to0\) everywhere and
\[ |f_n(x)|\le1. \]Since \(1\in L^1([0,1])\), Dominated Convergence gives
\[ \boxed{0}. \]Practice 8
We compute
\[ \|f_n\|_1 = \int_0^1\frac1n\,dx = \frac1n\to0. \]Hence
\[ \boxed{ f_n\to0 \text{ in }L^1 }. \]Practice 9
For every \(x>0\), eventually \(x>1/n\), so \(f_n(x)=0\). Thus \(f_n\to0\) almost everywhere.
But
\[ \|f_n\|_1 = n\left(\frac1n\right) = 1. \]Therefore,
\[ \boxed{ f_n\not\to0 \text{ in }L^1 }. \]Practice 10
Almost-everywhere convergence alone does not justify interchanging limit and integral. A theorem such as Monotone Convergence or Dominated Convergence must apply.
The sequence
\[ f_n=n\mathbf{1}_{(0,1/n)} \]converges to \(0\) almost everywhere, but
\[ \int_0^1f_n\,dx=1 \]for every \(n\). Therefore,
\[ \lim_{n\to\infty}\int_0^1f_n\,dx = 1 \ne 0 = \int_0^1\lim_{n\to\infty}f_n\,dx. \]What to Remember
Indicator functions satisfy
\[ \boxed{ \int_X\mathbf{1}_E\,d\mu = \mu(E) }. \]Nonnegative simple functions satisfy
\[ \boxed{ \int_X \sum_{k=1}^{n}a_k\mathbf{1}_{E_k}\,d\mu = \sum_{k=1}^{n}a_k\mu(E_k) }. \]The nonnegative integral is defined by approximation from below:
\[ \boxed{ \int_X f\,d\mu = \sup_{\substack{0\le\phi\le f\\\phi\text{ simple}}} \int_X\phi\,d\mu }. \]A measurable function belongs to \(L^1\) when
\[ \boxed{ \int_X|f|\,d\mu<\infty }. \]Monotone Convergence handles increasing nonnegative sequences. Fatou's Lemma gives one-sided limiting control. Dominated Convergence allows exact interchange of limit and integral under an integrable envelope.
The governing habit is: identify the convergence theorem, verify every hypothesis, and only then interchange the limit and integral.
Where We Go Next
Lebesgue integration naturally organizes functions by their integrability and convergence behavior. This leads to spaces such as \(L^1\), \(L^2\), and \(L^p\), alongside spaces of bounded, continuous, differentiable, and sequence-valued functions.
The next lesson, Function Spaces, develops these collections as mathematical objects equipped with norms, metrics, and completeness properties, preparing the transition into operators, linear functionals, and functional analysis.