Analysis & Linear Algebra

Lebesgue Measure and Integration

7 previous-year questions →every option and the correct answer, free Practice this topic →one question at a time, no clock, check as you go

See where the marks are across every paper →

1. Measurable sets and functions

Read this as a chapter →

Lec-11 Measurable sets

NPTEL · Measure and Integration

Lec-13 Characterization of Lebesgue measurable sets

NPTEL · Measure and Integration

Lec-14 Measurable functions

NPTEL · Measure and Integration

Measurability is preserved by every pointwise limit operation — the key advantage over Riemann.

Lebesgue measure and measurable functions free

In plain words

Small means three different things here: countable, measure zero, and containing no interval. No two of them imply each other, and the Cantor set is the standing warning about that.

Measure

Lebesgue measure m is countably additive on the $\sigma-$ algebra of measurable sets, translation invariant, and extends length.

Countable sets have measure 0. So $m(\mathbb{Q}) = 0$ .
Uncountable sets can have measure 0: the Cantor set.
Fat Cantor sets are nowhere dense with positive measure — so "small topologically" and "small measure-theoretically" are independent notions.
Non-measurable sets exist (Vitali) but need the axiom of choice.
Borel ⊊ Lebesgue measurable: the Lebesgue $\sigma-$ algebra is the completion of the Borel one.

Measurable functions

f is measurable ⇔ {x : f(x) > a} is measurable for every a.

Sums, products, sup, inf, limsup, liminf and pointwise limits of measurable functions are measurable — a much better closure property than continuity or Riemann integrability enjoy.
Continuous ⇒ Borel measurable ⇒ Lebesgue measurable.
Lusin: a measurable function is continuous off a set of arbitrarily small measure.
Egorov: on a finite-measure set, a.e. convergence is uniform off a set of arbitrarily small measure.

Almost everywhere

A property holds a.e. if the exceptional set has measure $0. \int$ |f| = 0 ⇔ f = 0 a.e. Changing f on a null set changes no integral.

Riemann vs Lebesgue

f bounded on [a,b] is Riemann integrable ⇔ its discontinuity set has measure 0; then it is Lebesgue integrable with the same value. $1_\mathbb{Q}$ is Lebesgue integrable (integral 0) but not Riemann integrable.

Improper Riemann integrals can exist without Lebesgue integrability: $\int_{0}^\infty \sin x/x$ converges but is not absolutely integrable, so it is not a Lebesgue integral.

Key takeaways

Measure zero ≠ countable (Cantor) and nowhere dense ≠ measure zero (fat Cantor).
Measurability survives all pointwise limit operations.
Riemann ⇒ Lebesgue for bounded functions on bounded intervals, but conditionally convergent improper integrals escape Lebesgue.

A set with no measure: the Vitali constructioninteractive

This visual is part of Trap Analysis

Why Lebesgue measure cannot be defined on every subset of ℝ — built in four steps.

Unlock interactive visuals

Exam focus: Measure zero, countable additivity and 'almost everywhere' are the workhorses. Know that measurable ⊋ Borel and that a non-measurable set requires the axiom of choice.

Practice

Part B2023 Decembershow ▾

Let X be a non-empty finite set and Y = {

f^{-1}(0)

: f is a real-valued function on X}. Which one of the following statements is true?

A.Y is an infinite set.
B.Y has $2^{|X|}$ elements.✓ Correct
C.There is a bijective function from X to Y.
D.There is a surjective function from X to Y.

Solution

Every subset of X is the zero set of its indicator-complement (f = 1 off the subset, 0 on it), so Y is the power set of X, with $2^{|X|}$ > |X| elements.

Save for spaced revision (free account)

Part C2023 Decembershow ▾

Let {

A_{n}

}

_{n\ge1}

be a collection of non-empty subsets of

\mathbb{Z}

such that

A_{n} \cap A_{m} = \emptyset

for m ≠ n. If

\mathbb{Z} = \cup_{n\ge1}

A_{n}

, then which of the following statements are necessarily true?

A. $A_{n}$ is finite for every integer n ≥ 1.
B. $A_{n}$ is finite for some integer n ≥ 1.
C. $A_{n}$ is infinite for some integer n ≥ 1.
D. $A_{n}$ is countable (finite or infinite) for every integer n ≥ 1.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2023 Juneshow ▾

Suppose S is an infinite set. Assuming that the axiom of choice holds, which of the following is true?

A.S is in bijection with the set of rational numbers.
B.S is in bijection with the set of real numbers.
C.S is in bijection with S × S.✓ Correct
D.S is in bijection with the power set of S.

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Let

E \subseteq \mathbb{R}

. Which of the following are true?

A.Every countable subset of $\mathbb{R}$ is Lebesgue measurable with measure zero.✓ Correct
B.If E is Lebesgue measurable, then E is a Borel set.
C.The Cantor set is uncountable but has Lebesgue measure zero.✓ Correct
D.A set of Lebesgue outer measure zero is measurable.✓ Correct

Solution

(1) a countable set is a countable union of singletons, each of outer measure zero.

(3) is the standard fact about the Cantor set.

(4) any set of outer measure zero satisfies the Carathéodory criterion automatically.

(2) is false: the Lebesgue $\sigma-$ algebra is the completion of the Borel $\sigma-$ algebra and is strictly larger — every subset of the (measure-zero) Cantor set is Lebesgue measurable, but there are 2^𝔠 of those and only 𝔠 Borel sets, so most of them are not Borel.

Save for spaced revision (free account)

Part Cshow ▾

Let

f : \mathbb{R} \to \mathbb{R}

. Which of the following are true?

A.If f is continuous, then f is Lebesgue measurable.✓ Correct
B.If f is monotone, then f is Lebesgue measurable.✓ Correct
C.The pointwise limit of a sequence of measurable functions, when it exists, is measurable.✓ Correct
D.Every Lebesgue measurable function is continuous at almost every point.

Solution

(1) continuous ⇒ Borel ⇒ Lebesgue measurable. (2) a monotone function has at most countably many discontinuities and its level sets are intervals — always measurable. (3) is a standard closure property (via limsup/liminf). (4) is false: the indicator of $\mathbb{Q} ($ Dirichlet's function) is Lebesgue measurable $(\mathbb{Q}$ is a countable, hence measurable, set) but is discontinuous at every real number — continuous at a set of measure zero, not almost every point.

Save for spaced revision (free account)

Part Cshow ▾

Let A ⊆ [0, 1] be Lebesgue measurable. Which of the following are true?

A.m(A) equals the supremum of m(K) over compact K ⊆ A (inner regularity).✓ Correct
B.m(A) equals the infimum of m(U) over open U ⊇ A (outer regularity).✓ Correct
C.If m(A) > 0, then A − A = {a − b : a, b ∈ A} contains an open interval around 0.✓ Correct
D.If m(A) > 0, then A must contain an open interval.

Solution

(1)–(2) are the regularity properties of Lebesgue measure. (3) is the Steinhaus theorem. (4) is false: a "fat Cantor set" — built like the usual Cantor set but removing shorter and shorter middle pieces — is nowhere dense (contains no interval at all) yet has positive measure.

Save for spaced revision (free account)

Part Bshow ▾

The Cantor middle-third set C satisfies which of the following?

A.C is countable with measure zero.
B.C is uncountable with measure zero.✓ Correct
C.C is uncountable with positive measure.
D.C is countable with positive measure.

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Let

(X, M, \mu)

be a measure space and

f, g : X \to \mathbb{R}

be measurable functions. Which of the following are true?

A.f + g is measurable.✓ Correct
B.max(f, g) is measurable.✓ Correct
C.If g = f almost everywhere and $\mu$ is complete, then g is measurable.✓ Correct
D.If |f| is measurable, then f is measurable.

Solution

(1) and (2) are standard closure properties of measurable functions.

(3) is exactly what completeness of $\mu$ buys you: a function equal a.e. to a measurable one differs from it only on a null set, and completeness guarantees every subset of a null set is measurable, which is what's needed to measure the points where f and g disagree.

(4) is false: let A ⊆ [0,1] be a non-Lebesgue-measurable set and set f = 1 on A, f = −1 off A.

Then |f| ≡ 1 is measurable, but f⁻ $^{1}($ {1}) = A is not.

Save for spaced revision (free account)

2. Lebesgue integral, MCT, DCT, Fatou

Read this as a chapter →

Lec-19 Monotone convergence theorem & Fatou's Lemma

NPTEL · Measure and Integration

MCT and Fatou, with the inequality direction made explicit.

Lec-20 Properties of Integral functions & Dominated Convergence Theorem

NPTEL · Measure and Integration

DCT — note exactly where the dominating function is used.

Lec-22 Lebesgue Integral and its properties

NPTEL · Measure and Integration

MCT, Fatou and dominated convergence free

In plain words

All three theorems answer one question: when may the limit go inside the integral? They exist because mass can escape sideways or upwards while the function tends to zero at every point.

The three theorems

Theorem	Hypothesis	Conclusion
Monotone convergence	$0 \le f_{1} \le f_{2} \le \cdots, f_{n} \to f a.e$ .	$\int{}f_{n} \to \int{}f$
Fatou	$f_{n} \ge 0$	$\int{}\liminf f_{n} \le \liminf \int{}f_{n}$
Dominated convergence	$f_{n} \to f a.e$ ., \| $f_{n}$ \| ≤ g with $\int{}g < \infty$	$\int{}f_{n} \to \int{}f$ and $\int$ \| $f_{n} - f$ \| → 0

Corollary of MCT: $\int\sum{}f_{n} = \sum\int{}f_{n}$ for non-negative $f_{n} ($ Beppo Levi / Tonelli).

Why domination is needed

Example on [0,1] or $\mathbb{R}$	$f_{n} \to 0$ pointwise	$\int{}f_{n}$
n· $1_{(0,1/n)}$	✓	1 — no convergence
$n^{2}\cdot1_{(0,1/n)}$	✓	$n \to \infty$
(1/n)· $1_{(0,n)}$	✓	1
$1_{(n, n+1)}$ (escaping bump)	✓	1

In each case no integrable g dominates the family, and Fatou's inequality is strict: $0 = \int{}\liminf < \liminf \int = 1$ .

Fubini and Tonelli

Tonelli: f ≥ 0 measurable ⇒ the iterated integrals agree (no integrability needed).
Fubini: f integrable on the product ⇒ iterated integrals agree.
Without one of these hypotheses the iterated integrals can differ: the classic $(x^{2} - y^{2})/(x^{2} + y^{2})^{2}$ on $[0,1]^{2}$ gives $\pi/4$ and $-\pi/4$ .

L^p and integration facts

$- \int$ |f| = 0 ⇔ f = 0 a.e.

f integrable ⇒ f finite a.e.
Absolute continuity of the integral: for $\varepsilon$ there is $\delta$ with $m(E) < \delta \Rightarrow \int_E$ |f| $< \varepsilon$ .

Key takeaways

Non-negative + increasing ⇒ MCT; non-negative alone ⇒ only Fatou's inequality.
DCT needs one integrable dominating function for the whole sequence.
The moving/shrinking bump is the universal counterexample when domination fails.

MCT, DCT and Fatou on one exampleinteractive

This visual is part of Trap Analysis

The same escaping-mass sequence run past all three theorems — which apply, which do not, and why.

Unlock interactive visuals

Exam focus: Pick the right convergence theorem: MCT (increasing, non-negative), Fatou (inequality, always), DCT (needs a dominating integrable g). The moving-bump examples show what happens without domination.

Practice

Part C2025 Juneshow ▾

Let

\mu

denote the Lebesgue measure on

\mathbb{R}

. Suppose that f is a non-negative Lebesgue measurable function on

\mathbb{R}

. Let

0 = a_{0} < a_{1} < a_{2} < \cdots

be an unbounded sequence such that

a_{n+1}

≤ ca

_{n}

for some real number c and for all n ≥ 1. Let A_k = {

x \in \mathbb{R}

| a_k ≤ f(x) <

a_{k+1}

} for each k ≥ 0. Which of the following statements are true?

A.If f is Lebesgue integrable on $\mathbb{R}$ , then $\sum_{k\ge0}$ $a_k \mu(A_k)$ is finite.✓ Correct
B.If $\sum_{k\ge0}$ $a_k \mu(A_k)$ is finite, then f is Lebesgue integrable on $\mathbb{R}$ .
C.If $\sum_{k\ge0}$ $a_k \mu(A_k)$ is finite, and $f(x) \ge a_{1}$ for all $x \in \mathbb{R}$ , then f is Lebesgue integrable on $\mathbb{R}$ .✓ Correct
D.If $\sum_{k\ge0}$ $a_k \mu(A_k)$ is finite and f is bounded, then f is Lebesgue integrable on $\mathbb{R}$ .

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Let

(f_{n})

be a sequence of measurable functions on a measure space

(X, \mu)

with

f_{n} \ge 0

. Which of the following are true?

A.By the Monotone Convergence Theorem, if $f_{n}$ ↑ f pointwise, then $\int{}f_{n} d\mu \to \int{}f d\mu$ .✓ Correct
B.By Fatou's Lemma, $\int \liminf f_{n} d\mu \le \liminf \int{}f_{n} d\mu$ .✓ Correct
C.By the Dominated Convergence Theorem, if $f_{n} \to f$ pointwise a.e. and | $f_{n}$ | ≤ g for an integrable g, then $\int{}f_{n} d\mu \to \int{}f d\mu$ .✓ Correct
D.In Fatou's Lemma, equality always holds — the inequality is never strict.

Solution

(1)–(3) are MCT, Fatou's Lemma, and DCT respectively.

(4) is false: the inequality in Fatou's Lemma can be strict.

Take $f_{n}$ to be the indicator of [n, n+1] on $\mathbb{R}$ — $f_{n} \to 0$ pointwise everywhere, so $\int{}\liminf f_{n} = 0$ , but $\int{}f_{n} = 1$ for every n, so $\liminf\int{}f_{n} = 1$ .

The mass escapes to infinity instead of vanishing.

Save for spaced revision (free account)

Part Cshow ▾

Let

f_{n} = n\cdot1_{(0, 1/n)}

on [0, 1] with Lebesgue measure. Which of the following are true?

A. $f_{n} \to 0$ pointwise everywhere on [0, 1].✓ Correct
B. $\int f_{n} \to 0$ .
C.The dominated convergence theorem applies to { $f_{n}$ }.
D.This example is consistent with Fatou's lemma: $\int \liminf f_{n} \le \liminf \int f_{n}$ .✓ Correct

Solution

For x > 0 the terms vanish once n > 1/x, and $f_{n}(0) = 0$ always, giving (1).

But $\int f_{n} = n\cdot(1/n) = 1$ for every n, so (2) fails.

(3) fails: a dominating function would need $g(x) \ge sup_{n} f_{n}(x) \approx 1/x$ near 0, which is not integrable — exactly why the limit of the integrals (1) differs from the integral of the limit (0).

(4) Fatou reads 0 ≤ 1, a strict inequality, which the lemma permits.

Save for spaced revision (free account)

Part Cshow ▾

Let

f_{n}

be the function on (0, 1) equal to n on the interval (0, 1/n) and 0 elsewhere, with Lebesgue measure. Which of the following are true?

A. $f_{n} \to 0$ pointwise on (0, 1).✓ Correct
B. $\int_{0}^{1} f_{n}$ dx = 1 for every n.✓ Correct
C.The Dominated Convergence Theorem applies to conclude $\int{}f_{n}$ dx $\to \int0$ dx = 0.
D. $\lim \int_{0}^{1} f_{n}$ dx $= \int_{0}^{1} (\lim f_{n})$ dx.

Solution

(1) for any fixed x > 0, once n > 1/x the spike has moved past x, so $f_{n}(x) = 0$ .

(2) the spike has height n and width 1/n, so its integral is n·(1/n) = 1 for every n.

(3) is false: there is no single integrable g dominating every $f_{n}$ , since the spike heights n are unbounded — DCT's hypothesis fails, which is exactly why the conclusion fails too.

(4) is false as a direct computation: the left side is lim 1 = 1, the right side is $\int0 = 0$ .

Save for spaced revision (free account)

Part Cshow ▾

Let

f_{n}(x) = n\cdot1_{(0, 1/n)}(x)

on [0,1]. Which of the following are true?

A. $f_{n} \to 0$ pointwise a.e.✓ Correct
B. $\int{}f_{n} \to 0$
C. $\liminf \int{}f_{n} > \int{}\liminf f_{n}$ ✓ Correct
D.The dominated convergence theorem applies

Solution

$\int{}f_{n} = 1$ for every n, so Fatou is strict and DCT cannot apply — no integrable g dominates all $f_{n}$ .

Save for spaced revision (free account)

3. L^p spaces essentials

Read this as a chapter →

Lec-34 Lp - spaces

NPTEL · Measure and Integration

Hölder, Minkowski and completeness (Riesz–Fischer).

Lec-35 L2(X,S,mue)

NPTEL · Measure and Integration

Why L² is the only Hilbert space in the family.

L^p spaces: inequalities and inclusions free

In plain words

Which of these spaces sits inside which depends entirely on what you are measuring. On a finite interval a bigger exponent gives a smaller space. For sequences it runs the other way.

The inequalities

Hölder: $\int$ |fg| ≤ ‖f‖_p‖g‖_q with 1/p + 1/q = 1. Cauchy–Schwarz is p = q = 2.
Minkowski: ‖f + g‖_p ≤ ‖f‖_p + ‖g‖_p (this is the triangle inequality).
Jensen: for a probability measure and convex $\varphi, \varphi(\int{}f) \le \int\varphi(f)$ .

Inclusions — opposite directions

Setting	Inclusion
finite measure (e.g. [0,1])	p < q ⇒ L^q ⊆ L^p
counting measure (sequences)	p < q ⇒ $\ell^p \subseteq \ell^q$
$\mathbb{R}$ with Lebesgue measure	no inclusion either way

Examples: $1/\sqrt{x} \in L^{1}[0,1]$ ∖ $L^{2}[0,1]; 1/x \in L^{2}[1,\infty)$ ∖ $L^{1}[1,\infty)$ ; the sequence 1/n is in $\ell^{2}$ but not $\ell^{1}$ .

Structure

L^p is a Banach space for $1 \le p \le \infty ($ Riesz–Fischer: complete). $- L^{2}$ is the only Hilbert space among them, with ⟨f,g⟩ $= \int{}f$ ḡ.
Dual: (L^p)* ≅ L^q for $1 \le p < \infty; (L^{1})$ * $= L^\infty$ but $(L^\infty)$ * ⊋ $L^{1}$ .
L^p is separable for $p < \infty$ ; $L^\infty$ is not.
For 0 < p < 1, ‖·‖_p is not a norm (triangle inequality fails).

Convergence relations

‖ $f_{n} - f$ ‖_p → 0 ⇒ a subsequence converges a.e., and $f_{n} \to f$ in measure. Neither implies the other in general: the typewriter sequence converges in L^p but nowhere a.e.; n· $1_{(0,1/n)}$ converges a.e. but not in $L^{1}$ .

Key takeaways

Finite measure: bigger p is a smaller space. Sequences: bigger p is a bigger space. $\mathbb{R}$ : neither. $- L^{2}$ is the Hilbert one; $L^\infty$ is the non-separable one.
L^p convergence gives an a.e. subsequence, never a.e. convergence itself.

Which Lᵖ contains which — and why it depends on the measureinteractive

This visual is part of Trap Analysis

The inclusion reverses between a probability space and the real line. Both directions, with the counterexample for each.

Unlock interactive visuals

Exam focus: L^p inclusions go one way on finite measure spaces and the other way for ℓ^p — getting the direction right is most of the battle.

Practice

Part C2025 Decembershow ▾

Let

\ell^{2}

denote the vector space of all square summable sequences {

a_{n}

}

_{n}\ge1

of real numbers with the inner product ⟨{

a_{n}

},{

b_{n}

}⟩

= \sum_{n}

₌

_{1}^\infty a_{n}b_{n}

. Let H = {{

a_{n}

}

\in\ell^{2}

: |

a_{n}

|≤1/n for all positive integers n}. Which of the following statements are true?

A.H contains an orthonormal basis of $\ell^{2}$ .
B.H is a linear subspace of $\ell^{2}$ .
C.H is a bounded subset of $\ell^{2}$ .✓ Correct
D.H is a convex subset of $\ell^{2}$ .✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

Let

\mu

denote the Lebesgue measure on

\mathbb{R}

and

f : \mathbb{R} \to [0, \infty)

be a Lebesgue measurable function. For n ≥ 1, let

E_{n} =

{

x \in \mathbb{R} : n - 1 \le f(x) < n

}. Suppose that

\sum(n\ge1) n^{2}\mu(E_{n}) < \infty

. Which of the following statements are true?

A. $f \in L^{1}(\mathbb{R})$ ✓ Correct
B. $f \in L^{2}(\mathbb{R})$ ✓ Correct
C. $f \in L^{4}(\mathbb{R})$
D. $f \in L^\infty(\mathbb{R})$

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

Consider the real vector space X = {

f : [0, 1] \to \mathbb{R}

| f is continuous}, along with the norms ‖·‖

_{1}

and ‖·‖

_{2}

defined by ‖f‖

_{1} = \int_{0}^{1}

|f(x)|dx and ‖f‖

_{2} = (\int_{0}^{1}

|f(x)|

^{2}

dx)^(1/2). For n ≥ 1 and x ∈ [0, 1], let

f_{n}(x) =

^{n}

. Which of the following statements are true?

A.(‖ $f_{n}$ ‖ $_{1})$ is a convergent sequence.✓ Correct
B.(‖ $f_{n}$ ‖ $_{2})$ is a convergent sequence.
C.Both (‖ $f_{n}$ ‖ $_{1})$ and (‖ $f_{n}$ ‖ $_{2})$ are convergent sequences.
D.Neither (‖ $f_{n}$ ‖ $_{1})$ nor (‖ $f_{n}$ ‖ $_{2})$ is a convergent sequence.

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Which of the following are true?

A. $L^{2}[0,1] \subseteq L^{1}[0,1]$ ✓ Correct
B. $L^{1}[0,1] \subseteq L^{2}[0,1]$
C. $\ell^{1} \subseteq \ell^{2}$ ✓ Correct
D. $L^{1}(\mathbb{R}) \subseteq L^{2}(\mathbb{R})$

Solution

Finite measure reverses the inclusion (Hölder); for sequences the inclusion goes up.

On $\mathbb{R}$ neither holds: $1/\sqrt{x}\cdot1_{(0,1)}$ is in $L^{1}$ not $L^{2}$ , and 1/x· $1_{(1,\infty)}$ is in $L^{2}$ not $L^{1}$ .

Save for spaced revision (free account)

Part Cshow ▾

Let

(X, \mu)

be a measure space with

\mu(X) = 1 (a

probability space). Which of the following are true?

A. $L^{2}(\mu) \subseteq L^{1}(\mu)$ .✓ Correct
B. $L^{1}(\mu) \subseteq L^{2}(\mu)$ .
C.For $f \in L^{2}(\mu)$ , ‖f‖ $_{1} \le$ ‖f‖ $_{2}$ .✓ Correct
D.On $\mathbb{R}$ with Lebesgue measure, $L^{2} \subseteq L^{1}$ also holds.

The worked solution and the per-option analysis are in Trap Analysis.