Integration and Differentiation of Measures
Y. Eddie Lu, Summer 2026
ECON 8002 course index · Lecture 2 of 9
Orientation
How to use this page
First identify the measure. Then identify whether the integrand is nonnegative, integrable, or merely defined almost everywhere. Never exchange a limit, derivative, or integral without naming the theorem that permits it.
How should I think about this lecture?
Simple payoffs
Null sets and limits
\(\longrightarrow\)
Lebesgue integration makes expectation and density rigorous
\(\longrightarrow\)
Interchange theorems
RN derivatives and PDFs
Course map
- Measure spaces
- Integration and Radon–Nikodym derivatives (current lecture)
- Distribution properties
- Conditional expectation
- Convergence modes
- Continuous mapping and Slutsky
- Laws of large numbers
- Weak limits and the delta method
- Central limit theorems and inference
Prerequisite retrieval
From Lecture 1: \(X\) is measurable, \(P_X(B)=P(X\in B)\), and a nonnegative measurable function is approximated from below by simple functions.
Retrieval check.
If \(s=2\mathbf1_A+5\mathbf1_{A^c}\) and \(P(A)=0.3\), compute \(\int s\,dP\).
\(2(0.3)+5(0.7)=4.1\).
Constructing the integral
Motivation: pointwise convergence is insufficient
On \([0,1]\), let \(f_n(x)=n\mathbf1_{(0,1/n)}(x)\). Then \(f_n(x)\to0\) for every \(x\), but \[\int_0^1f_n(x)\,dx=1\not\to0.\] The integral and limit cannot be swapped merely because convergence is pointwise.
Construction ladder
Define the integral for nonnegative simple functions. Extend it by approximation to nonnegative measurable functions. Then write a signed function as \(f=f^+-f^-\).
Definition: simple-function integral
If \(s=\sum_{j=1}^ma_j\mathbf1_{A_j}\) with \(a_j\ge0\) and \(A_j\in\mathcal F\), define \[\int s\,d\mu=\sum_{j=1}^ma_j\mu(A_j).\]
This value does not depend on the particular simple-function representation.
Nonnegative measurable functions
For measurable \(f\ge0\), \[\int f\,d\mu=\sup\left\{\int s\,d\mu:0\le s\le f,\ s\text{ simple}\right\}.\] The value may be \(+\infty\).
Retrieval check.
Why is the definition a supremum rather than an arbitrary limit of approximations?
It is representation-free: it collects every admissible simple lower approximation and selects their least upper bound.
Signed functions and integrability
Define \(f^+=\max(f,0)\) and \(f^-=\max(-f,0)\). Then \(f=f^+-f^-\) and \(|f|=f^++f^-\). The signed integral exists when at least one of \(\int f^+d\mu,\int f^-d\mu\) is finite. It is integrable when both are finite, equivalently \[\int|f|\,d\mu<\infty.\]
Expectations are integrals
On \((\Omega,\mathcal F,P)\), \[E[X]=\int_\Omega X\,dP=\int_{\mathbb R}x\,dP_X(x),\] when \(X\) is integrable. With counting measure this is a sum; with Lebesgue density \(p\) it is \(\int xp(x)dx\).
Linearity and a.e. equality
If the relevant integrals exist and one summand is integrable, \[\int(af+bg)\,d\mu=a\int f\,d\mu+b\int g\,d\mu.\] “\(f=g\) almost everywhere” means \(\mu\{f\ne g\}=0\). Integrals ignore this set, but pointwise statements do not.
Order and integral bounds: Proposition 2.3 and Corollary 2.4
For measurable \(f,g\) whose integrals exist, \(f\le g\) a.e. implies \(\int f\,d\mu\le\int g\,d\mu\). If \(f\ge0\) a.e. and \(\int f\,d\mu=0\), then \(f=0\) a.e. If \(f\) is integrable, \[\left|\int f\,d\mu\right|\le\int|f|\,d\mu.\] Finally, if \(f=g\) a.e., their integrals agree whenever defined.
Retrieval check.
Can a function be nonzero yet have integral \(0\)?
Yes, if it is nonzero only on a null set. For \(f\ge0\), \(\int f\,d\mu=0\) implies \(f=0\) a.e.
Limit and interchange theorems
\(\liminf\) and \(\limsup\)
For an extended-real sequence \((a_n)\), \[\liminf_na_n=\sup_N\inf_{n\ge N}a_n,\qquad \limsup_na_n=\inf_N\sup_{n\ge N}a_n.\] The ordinary finite limit exists exactly when these two quantities agree finitely. For functions, apply the definitions pointwise.
The three interchange theorems
| Theorem | Assumptions | Conclusion |
|---|---|---|
| Fatou | measurable \(f_n\ge0\) | \(\int\liminf f_n\le\liminf\int f_n\) |
| DCT | measurable \(f_n\); \(f_n\to f\) a.e.; \(|f_n|\le g\) a.e.; measurable \(g\in L^1\) | \(\int f_n\to\int f\) |
| MCT | measurable \(0\le f_n\uparrow f\) | \(\int f_n\uparrow\int f\) |
Fatou: the safe direction
Fatou is one-sided. It handles nonnegative terms and says the integral of the eventual lower behavior cannot exceed the eventual lower bound of integrals. It does not state equality.
Dominated convergence
If measurable \(f_n\) satisfy \(f_n\to f\) a.e. and one measurable, integrable \(g\) satisfies \(|f_n|\le g\) a.e. for every \(n\), then \(f\) is measurable and integrable and \[\lim_{n\to\infty}\int f_n\,d\mu=\int f\,d\mu.\]
DCT proof map
- From \(|f_n|\le g\), both \(g+f_n\) and \(g-f_n\) are nonnegative.
- Apply Fatou to \(g+f_n\) to bound \(\int f\) above by \(\liminf\int f_n\).
- Apply Fatou to \(g-f_n\) to bound \(\limsup\int f_n\) above by \(\int f\).
- The bounds meet.
Retrieval check.
For \(f_n(x)=x^n\) on \([0,1]\), name an integrable dominator.
\(g(x)=1\). Thus DCT gives \(\int_0^1x^n dx\to0\).
Monotone convergence
If measurable \(f_n\) satisfy \(0\le f_1\le f_2\le\cdots\) and \(f_n\to f\) a.e., then \[\int f_n\,d\mu\uparrow\int f\,d\mu,\] including the possible value \(+\infty\).
Differentiation under the integral sign
Fix an open interval \(I\). Assume \(h(\cdot,t)\) is measurable for every \(t\in I\), \(h(\cdot,t_*)\) is integrable for some \(t_*\in I\), and there is one measurable \(N\) with \(\mu(N)=0\) such that, for every \(\omega\notin N\), \(t\mapsto h(\omega,t)\) is differentiable throughout \(I\). Assume one \(g\in L^1(\mu)\) satisfies \[|\partial_t h(\omega,t)|\le g(\omega)\quad\text{for every }t\in I,\ \omega\notin N.\] Then \(H(t)=\int h(\omega,t)\,d\mu\) is finite and differentiable on \(I\), with \[H'(t)=\int\partial_th(\omega,t)\,d\mu.\]
Why DCT proves it
The difference quotient converges pointwise to \(\partial_th\). The mean-value theorem bounds each quotient by \(g\). DCT then passes its limit through the integral.
Retrieval check.
What is the role of the mean-value theorem here?
It turns a difference quotient into a derivative evaluated at an intermediate point, so the derivative bound supplies the DCT dominator.
Transformations and iterated integration
Change of variables: the measure-theoretic version
For measurable \(T:(\Omega,\mathcal F)\to(S,\mathcal G)\) and measurable \(g:S\to\mathbb R\), \[\int_\Omega g(T(\omega))\,d\mu(\omega)=\int_Sg(s)\,d(\mu\circ T^{-1})(s),\] whenever either side exists.
Proof architecture for change of variables
Prove it for \(g=\mathbf1_B\), extend by linearity to nonnegative simple \(g\), use MCT for nonnegative measurable \(g\), and subtract positive/negative parts for signed \(g\). This is why Lecture 1 built simple approximations.
Expectation depends on the law
Taking \(T=X\) and \(\mu=P\) gives \[E[g(X)]=\int_{\mathbb R}g(x)\,dP_X(x).\] This identity needs no Lebesgue density and is the robust form of “integrate using the distribution of \(X\).”
Retrieval check.
Does change of variables require \(T\) to be one-to-one?
No. It uses the induced measure \(\mu\circ T^{-1}\), not an inverse map or Jacobian formula.
Fubini: iterated integration
For \(\sigma\)-finite \(\mu_1,\mu_2\) and integrable \(f\) on the product space, \[\int f\,d(\mu_1\otimes\mu_2)=\int\left[\int f(x,y)\,d\mu_1(x)\right]d\mu_2(y),\] and the reverse order agrees.
Economic use: heterogeneous agents
Fubini justifies integrating a payoff first over idiosyncratic shocks and then over types, when the integrability conditions hold. It is the formal basis for many iterated-expectation manipulations.
Nonexample: an unjustified order swap
If \(f\) is not integrable, neither iterated integral need be finite or equal to a joint integral. The habit is: first establish nonnegativity for Tonelli-type reasoning or absolute integrability for Fubini, then change the order.
Radon–Nikodym differentiation
Radon–Nikodym motivation
Suppose a measure \(\nu\) is formed by weighting another measure \(\mu\): \[\nu(A)=\int_Af\,d\mu.\] Then every \(\mu\)-null set is \(\nu\)-null. The RN theorem nearly reverses this implication.
Lemma 2.8: integrating over sets creates a measure
If \(f:\Omega\to[0,\infty]\) is measurable, then \[\nu(A)=\int_A f\,d\mu,\qquad A\in\mathcal F,\] defines a measure on \((\Omega,\mathcal F)\) and \(\nu\ll\mu\).
Absolute continuity and RN theorem
\(\nu\ll\mu\) means \(\mu(A)=0\Rightarrow\nu(A)=0\) for every measurable \(A\).
If \(\mu\) is \(\sigma\)-finite and \(\nu\ll\mu\), then there is measurable \(f\ge0\), unique \(\mu\)-a.e., such that \[\nu(A)=\int_Af\,d\mu.\qquad f=\frac{d\nu}{d\mu}\]
Corollary 2.10: uniqueness from set integrals
If an integrable measurable \(h\) satisfies \[\int_Ah\,d\mu=0\quad\text{for every }A\in\mathcal F,\] then \(h=0\) \(\mu\)-a.e. Apply this to the difference of two candidate densities to obtain RN uniqueness.
Retrieval check.
Is \(\nu\ll\mu\) symmetric?
No. It says only that \(\mu\)-null sets are also \(\nu\)-null. Mutual absolute continuity requires both directions.
Densities depend on a reference measure
A PMF is \(dP/d\#\) relative to counting measure. An ordinary PDF is \(dP/d\lambda\) relative to Lebesgue measure. A mixed distribution may lack a Lebesgue density yet have a density relative to a measure such as \(\lambda+\delta_c\).
RN calculus: substitution and additivity
Let \(\mu\) be \(\sigma\)-finite. If \(\nu\ll\mu\) and measurable \(h\ge0\) or \(h\in L^1(\nu)\), then \[\int h\,d\nu=\int h\frac{d\nu}{d\mu}\,d\mu.\] If \(\nu_1,\nu_2\ll\mu\), then \[\frac{d(\nu_1+\nu_2)}{d\mu}=\frac{d\nu_1}{d\mu}+\frac{d\nu_2}{d\mu}\quad\mu\text{-a.e.}\]
RN calculus: chain, reciprocal, and product
If \(\rho\ll\nu\ll\mu\) and \(\mu,\nu\) are \(\sigma\)-finite, then \[\frac{d\rho}{d\mu}=\frac{d\rho}{d\nu}\frac{d\nu}{d\mu}\quad\mu\text{-a.e.}\] If also \(\mu\ll\nu\), then \(d\nu/d\mu=(d\mu/d\nu)^{-1}\) almost everywhere with respect to either measure. For \(\sigma\)-finite \(\nu_j\ll\mu_j\) with each \(\mu_j\) also \(\sigma\)-finite, \[\frac{d(\nu_1\otimes\nu_2)}{d(\mu_1\otimes\mu_2)}(x,y) =\frac{d\nu_1}{d\mu_1}(x)\frac{d\nu_2}{d\mu_2}(y)\] for \((\mu_1\otimes\mu_2)\)-a.e. \((x,y)\).
Synthesis and retrieval
Assumption audit
| Assumption | What it buys |
|---|---|
| \(f_n\to f\) a.e. | a limiting integrand |
| one \(g\in L^1\) dominates | DCT interchange |
| \(0\le f_n\uparrow f\) | MCT interchange without dominator |
| integrability of \(f\) on product | Fubini order swap |
| \(\mu\) \(\sigma\)-finite, \(\nu\ll\mu\) | RN density |
Common errors
“Each \(f_n\) is integrable” is not DCT. A.e. equality is not pointwise equality. \(\nu\ll\mu\) is directional. A density is not automatically a derivative with respect to Lebesgue measure.
Calculation blueprint
- State the base measure and domain. 2. Check sign/integrability. 3. For an interchange, identify Fatou, DCT, MCT, or Fubini and verify every assumption. 4. For a transformation, use the pushforward law. 5. For a density, name its reference measure.
Exit ticket 1
Retrieval check.
State DCT with every assumption.
If \(f_n\to f\) a.e. and \(|f_n|\le g\) a.e. for one integrable \(g\), then \(\int f_n d\mu\to\int f d\mu\).
Exit ticket 2
Retrieval check.
Which theorem applies to \(0\le f_n\uparrow f\) when no integrable dominator is known?
MCT. Its monotonicity and nonnegativity replace domination.
Exit ticket 3
Retrieval check.
Why can a discrete PMF be called a density?
It is the RN derivative of its probability measure with respect to counting measure.
Mastery checklist
You should now be able to construct an integral from simple functions, distinguish DCT/Fatou/MCT, justify a change of variables, state Fubini’s conditions, and define an RN derivative relative to its reference measure.
Built from ECON 8002 Lecture 2. The source’s symbols are normalized here: \(\mu,\nu\) denote measures and \(f,g\) functions.