Which statements about Lebesgue measure on ℝ are true?

every countable subset of ℝ has measure 0

the Cantor ternary set is uncountable and has measure 0

some closed nowhere dense subset of [0, 1] has positive measure

every countable subset of ℝ has measure 0

every subset of ℝ of measure 0 is countable

the Cantor ternary set is uncountable and has measure 0

some closed nowhere dense subset of [0, 1] has positive measure

Let f = 1_{ℚ} on [0, 1] (1 at rationals, 0 at irrationals). Then

f is not Riemann integrable, and its Lebesgue integral is 0

f is Riemann integrable with integral 0

f is Lebesgue integrable with integral 1

Which of the following bounded functions on [0, 1] are Riemann integrable?

the indicator function of the Cantor ternary set

Thomae’s function: 1/q at x = p/q in lowest terms, 0 at irrationals

sin(1/x) for x ≠ 0, with value 0 at x = 0

the indicator function of the Cantor ternary set

Thomae’s function: 1/q at x = p/q in lowest terms, 0 at irrationals

the indicator function of a fat Cantor set of measure 1/2

sin(1/x) for x ≠ 0, with value 0 at x = 0

For fₙ = n·1_{(0,1/n)} on (0, 1), which statement is correct?

fₙ → 0 at every point and ∫fₙ = 1 for all n, so the inequality in Fatou’s lemma is strict

the dominated convergence theorem gives lim ∫fₙ = 0

the monotone convergence theorem gives lim ∫fₙ = 0

fₙ does not converge at points close to 0

Which statements are true?

fₙ = 1_{[n, n+1]} converges to 0 at every point of ℝ while ∫fₙ = 1 for every n

the monotone convergence theorem holds even when the limit function has infinite integral

fₙ = 1_{[n, n+1]} converges to 0 at every point of ℝ while ∫fₙ = 1 for every n

for the sequence in the first statement some integrable g satisfies |fₙ| ≤ g for all n

the monotone convergence theorem holds even when the limit function has infinite integral

if fₙ → f a.e. and ∫|fₙ| ≤ 1 for all n, then ∫fₙ → ∫f

Which statements about Lᵖ spaces (Lebesgue measure) are true?

f(x) = 1/x belongs to L²(1, ∞) but not to L¹(1, ∞)

Which statements about measurable functions on ℝ (Lebesgue measure) are true?

the pointwise limit of a sequence of measurable functions is measurable

every continuous function f: ℝ → ℝ is measurable

if f is measurable and g = f almost everywhere, then g is measurable

the pointwise limit of a sequence of measurable functions is measurable

every continuous function f: ℝ → ℝ is measurable

if f is measurable and g = f almost everywhere, then g is measurable

For 1 < p < ∞ and q with 1/p + 1/q = 1, which statement about Lᵖ[0, 1] is correct?

its dual space is isometrically isomorphic to L^{q}[0, 1]

its dual is L^{p}[0, 1] for every p

the continuous functions are not dense in it

Lebesgue Measure and Integration, the Convergence Theorems and Lp Spaces

The last part of Section 3 is Lebesgue theory: Lebesgue measure on the real line, measurable functions, the Lebesgue integral, the three convergence theorems — Fatou’s lemma, the monotone convergence theorem and the dominated convergence theorem — and the Lᵖ spaces. Riemann integrability is treated here as well, through Lebesgue’s criterion, because the comparison between the two integrals is where most questions start. GATE sets this topic in two ways: as a limit of integrals to be evaluated by exchanging limit and integral, where the job is to name the theorem and the dominating function; and as statements about null sets, measurability and Lᵖ inclusions, where the job is to know the Cantor set, the fat Cantor set, the sliding bump and the functions x^−1/2 and 1/x.

1. Lebesgue measure on the real line

The outer measure of E ⊆ ℝ is m∗(E) = inf Σℓ(Iₖ) over countable covers of E by open intervals. The Lebesgue measurable sets form a σ-algebra containing all Borel sets, on which m = m∗ is countably additive and translation invariant, with m([a, b]) = b − a. Continuity: m(⋃Eₖ) = lim m(Eₖ) for increasing sets, and m(⋂Eₖ) = lim m(Eₖ) for decreasing sets if m(E₁) < ∞ (the sets [k, ∞) decrease to ∅ with infinite measure throughout).

Null sets and their surprises
Set	Measure	What it shows
any countable set, e.g. ℚ	0	cover the k-th point by an interval of length ε/2ᵏ
[0, 1] ∖ ℚ	1	a dense set of full measure with empty interior
the Cantor set	0	uncountable yet null: 1 − Σ 2ᵏ⁻¹/3ᵏ = 0
fat Cantor set: at step k remove 2ᵏ⁻¹ middle intervals of length 4⁻ᵏ	1 − Σ 2ᵏ⁻¹/4ᵏ = 1/2	closed, nowhere dense, of positive measure

ℹ️ Non-measurable sets exist, but only by choice

Vitali’s set picks one point from each coset of ℚ in [0, 1] (using the axiom of choice). Its rational translates by ℚ ∩ [−1, 1] are disjoint, lie in [−1, 2] and cover [0, 1]; countable additivity would give 1 ≤ Σm(V) ≤ 3, impossible whether m(V) is 0 or positive. So not every subset of ℝ is Lebesgue measurable.

2. Measurable functions and the Lebesgue integral

f: E → [−∞, ∞] is measurable if {f > a} is measurable for every real a. Continuous functions, monotone functions, sums, products, sup, inf, limsup, liminf and pointwise limits of measurable functions are measurable, and because Lebesgue measure is complete, a function equal almost everywhere (a.e.) to a measurable one is measurable. Egorov: on a set of finite measure, a.e. convergence is uniform off a set of arbitrarily small measure.

The integral of a non-negative simple function Σcₖ1_Aₖ is Σcₖm(Aₖ); for f ≥ 0 measurable, ∫f is the supremum over simple 0 ≤ s ≤ f; f is integrable if ∫|f| < ∞, and then ∫f = ∫f⁺ − ∫f⁻. Changing f on a null set does not change ∫f, and ∫|f| = 0 iff f = 0 a.e.

Riemann against Lebesgue
Function	Riemann	Lebesgue
bounded f on [a, b], continuous except on a null set	integrable (Lebesgue’s criterion: iff the discontinuities form a null set)	integrable, same value
1_ℚ on [0, 1] (Dirichlet)	not integrable: upper sums 1, lower sums 0	∫ = 0
Thomae’s function; 1 on the Cantor set	integrable, value 0 (discontinuities ℚ, resp. the Cantor set, are null)	∫ = 0
sin x/x on (0, ∞)	improper integral converges to π/2	not integrable: ∫\|sin x/x\| = ∞

3. Fatou, monotone convergence and dominated convergence

Monotone convergence theorem (MCT). If 0 ≤ f₁ ≤ f₂ ≤ … are measurable and fₙ → f pointwise, then ∫fₙ → ∫f (both sides may be ∞). Example: (1 − x/n)ⁿ1_[0,n] increases to e^−x, so ∫₀ⁿ (1 − x/n)ⁿ dx = n/(n + 1) → ∫₀^∞ e^−x dx = 1.
Fatou’s lemma. For measurable fₙ ≥ 0, ∫ liminf fₙ ≤ liminf ∫fₙ. The inequality can be strict: fₙ = n·1_(0,1/n) tends to 0 everywhere while ∫fₙ = 1.
Dominated convergence theorem (DCT). If fₙ → f a.e. and |fₙ| ≤ g for an integrable g, then f is integrable and ∫|fₙ − f| → 0, so ∫fₙ → ∫f. Example: |n sin(x/n)/(x(1 + x²))| ≤ 1/(1 + x²) because |sin t| ≤ |t|, so the integral over (0, ∞) tends to ∫₀^∞ dx/(1 + x²) = π/2.

⚠️ The sliding bump has no dominating function

fₙ = 1_{[n, n+1]} tends to 0 at every point of ℝ while ∫fₙ = 1 for every n. DCT does not apply because sup fₙ = 1_[1,∞) is not integrable, and MCT does not apply because the sequence is not increasing. A uniform bound ∫|fₙ| ≤ 1 is not domination either: n·1_(0,1/n) has it and still loses its mass in the limit.

4. The Lp spaces

For 1 ≤ p < ∞, Lᵖ(E) is the space of measurable f with ‖f‖ₚ = (∫|f|ᵖ)^1/p < ∞, functions equal a.e. being identified; L^∞ uses the essential supremum. Hölder: ‖fg‖₁ ≤ ‖f‖ₚ‖g‖_q with 1/p + 1/q = 1; Minkowski: ‖f + g‖ₚ ≤ ‖f‖ₚ + ‖g‖ₚ. Riesz–Fischer: every Lᵖ, 1 ≤ p ≤ ∞, is complete, a Banach space, and L² is a Hilbert space with ⟨f, g⟩ = ∫fḡ. For 1 ≤ p < ∞, Lᵖ is separable, continuous functions are dense in it and its dual is L^q; L^∞ is not separable, and the dual of L^∞ is strictly bigger than L¹.

Inclusions depend on the measure of the space
Setting	Inclusion	Witness against the reverse
finite measure, e.g. [0, 1]	L^q ⊆ Lᵖ for p < q, with ‖f‖ₚ ≤ m(E)^{1/p − 1/q}‖f‖_q	x^−1/2 ∈ L¹(0, 1) ∖ L²(0, 1)
(1, ∞)	no inclusion either way in general	1/x ∈ L²(1, ∞) ∖ L¹(1, ∞)
sequence spaces ℓᵖ (counting measure)	ℓᵖ ⊆ ℓ^q for p < q, the reverse of [0, 1]	(1/n) ∈ ℓ² ∖ ℓ¹

🧠 Power functions decide membership

x^−a is in Lᵖ(0, 1) iff ap < 1, and in Lᵖ(1, ∞) iff ap > 1. So x^−1/3 ∈ Lᵖ(0, 1) exactly for p < 3, and the supremum of such p is 3 without being attained. Near 0 small powers are safe; near ∞ large powers are.

Key takeaways

Countable sets and the Cantor set are null; a fat Cantor set is nowhere dense with positive measure; non-measurable sets need the axiom of choice.
A bounded function on [a, b] is Riemann integrable iff its discontinuities form a null set, and then the two integrals agree.
MCT for increasing non-negative sequences, DCT when an integrable g dominates, Fatou as an inequality for everything non-negative.
The sliding bump and n·1_(0,1/n) are the counterexamples that show why domination is needed.
Lᵖ is a Banach space; on finite measure spaces higher p is smaller, on ℓᵖ the reverse, and on (0, ∞) neither.

Practice questions (14)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The Lebesgue measure of the set of irrational numbers in [0, 2] is ____.
Numerical answer — type the value.
Show answer
Answer: 2
ℚ ∩ [0, 2] is countable, hence of measure 0, so m([0, 2] ∖ ℚ) = m([0, 2]) − 0 = 2. The irrationals have empty interior and are still of full measure, which is why measure and topology must be kept apart.
Which statements about Lebesgue measure on ℝ are true?
1. every countable subset of ℝ has measure 0
2. every subset of ℝ of measure 0 is countable
3. the Cantor ternary set is uncountable and has measure 0
4. some closed nowhere dense subset of [0, 1] has positive measure
Show answer
Answer: A — every countable subset of ℝ has measure 0; C — the Cantor ternary set is uncountable and has measure 0; D — some closed nowhere dense subset of [0, 1] has positive measure
(1) Cover the k-th point by an interval of length ε/2ᵏ. (3) The Cantor set is in bijection with 0–2 ternary expansions, hence uncountable, and the removed intervals have total length Σ2ᵏ⁻¹/3ᵏ = 1. (4) A fat Cantor set works. (2) is false: the Cantor set is the counterexample.
From [0, 1] remove the open middle interval of length 1/4; from each of the two remaining intervals remove the open middle interval of length 1/16; at step k remove from each of the 2ᵏ⁻¹ remaining intervals the open middle interval of length 4⁻ᵏ. The Lebesgue measure of the set that remains is ____.
Numerical answer — type the value.
Show answer
Answer: 0.5
The removed length is Σ_k≥1 2ᵏ⁻¹ · 4⁻ᵏ = (1/4) Σ_k≥1 (1/2)ᵏ⁻¹ = (1/4) · 2 = 1/2, so the remainder has measure 1 − 1/2 = 0.5. It is closed and contains no interval, so it is a nowhere dense set of positive measure. The ternary Cantor construction removes total length 1.
Let f = 1_ℚ on [0, 1] (1 at rationals, 0 at irrationals). Then
1. f is not Riemann integrable, and its Lebesgue integral is 0
2. f is Riemann integrable with integral 0
3. f is not Lebesgue measurable
4. f is Lebesgue integrable with integral 1
Show answer
Answer: A — f is not Riemann integrable, and its Lebesgue integral is 0
Every subinterval contains rationals and irrationals, so every upper Darboux sum is 1 and every lower sum is 0: not Riemann integrable (f is discontinuous everywhere). But f = 0 a.e. because ℚ is null and ℚ is a Borel set, so f is measurable and ∫f = m(ℚ ∩ [0, 1]) = 0.
Which of the following bounded functions on [0, 1] are Riemann integrable?
1. the indicator function of the Cantor ternary set
2. Thomae’s function: 1/q at x = p/q in lowest terms, 0 at irrationals
3. the indicator function of a fat Cantor set of measure 1/2
4. sin(1/x) for x ≠ 0, with value 0 at x = 0
Show answer
Answer: A — the indicator function of the Cantor ternary set; B — Thomae’s function: 1/q at x = p/q in lowest terms, 0 at irrationals; D — sin(1/x) for x ≠ 0, with value 0 at x = 0
Use Lebesgue’s criterion. (1) The indicator of a closed set C is discontinuous exactly on the boundary of C, here C itself, which is null. (2) Thomae’s function is discontinuous exactly at the rationals, a null set. (3) The fat Cantor set is its own boundary, of measure 1/2 > 0: not integrable. (4) The only discontinuity is x = 0.
The value of lim_n→∞ ∫₀ⁿ (1 − x/n)ⁿ dx is ____.
Numerical answer — type the value.
Show answer
Answer: 1
The functions (1 − x/n)ⁿ1_[0,n](x) increase with n to e^−x (since log(1 − t) ≤ −t and the sequence is increasing), so by the monotone convergence theorem the limit is ∫₀^∞ e^−x dx = 1. Directly, ∫₀ⁿ (1 − x/n)ⁿ dx = n/(n + 1) → 1.
The value of lim_n→∞ ∫₀¹ (1 + x/n)⁻ⁿ dx, correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.63
(1 + x/n)⁻ⁿ → e^−x pointwise and 0 < (1 + x/n)⁻ⁿ ≤ 1, an integrable bound on [0, 1], so by dominated convergence the limit is ∫₀¹ e^−x dx = 1 − 1/e = 0.632, i.e. 0.63.
The value of lim_n→∞ ∫₀^∞ n sin(x/n) / (x(1 + x²)) dx, correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 1.57
For fixed x > 0, n sin(x/n)/x → 1, so the integrand tends to 1/(1 + x²). Since |sin t| ≤ |t|, |n sin(x/n)/x| ≤ 1 and the integrand is dominated by 1/(1 + x²), integrable on (0, ∞). DCT gives ∫₀^∞ dx/(1 + x²) = π/2 = 1.5708, i.e. 1.57.
For fₙ = n·1_(0,1/n) on (0, 1), which statement is correct?
1. fₙ → 0 at every point and ∫fₙ = 1 for all n, so the inequality in Fatou’s lemma is strict
2. the dominated convergence theorem gives lim ∫fₙ = 0
3. the monotone convergence theorem gives lim ∫fₙ = 0
4. fₙ does not converge at points close to 0
Show answer
Answer: A — fₙ → 0 at every point and ∫fₙ = 1 for all n, so the inequality in Fatou’s lemma is strict
For each x ∈ (0, 1), fₙ(x) = 0 once n > 1/x, so the limit is 0 everywhere, while ∫fₙ = n · (1/n) = 1. Fatou gives 0 ≤ 1, strictly. DCT fails (sup fₙ ≥ 1/(2x) is not integrable) and MCT fails (the sequence is not increasing).
Which statements are true?
1. fₙ = 1_{[n, n+1]} converges to 0 at every point of ℝ while ∫fₙ = 1 for every n
2. for the sequence in the first statement some integrable g satisfies |fₙ| ≤ g for all n
3. the monotone convergence theorem holds even when the limit function has infinite integral
4. if fₙ → f a.e. and ∫|fₙ| ≤ 1 for all n, then ∫fₙ → ∫f
Show answer
Answer: A — fₙ = 1_{[n, n+1]} converges to 0 at every point of ℝ while ∫fₙ = 1 for every n; C — the monotone convergence theorem holds even when the limit function has infinite integral
(1) is the sliding bump. (2) Any such g would be ≥ 1 on [1, ∞), so not integrable. (3) MCT is a statement in [0, ∞]: e.g. fₙ = min(n, 1/x) on (0, 1) increases to 1/x and ∫fₙ → ∞. (4) is false: n·1_(0,1/n) has ∫ = 1 and limit 0.
Which statements about Lᵖ spaces (Lebesgue measure) are true?
1. L²[0, 1] ⊆ L¹[0, 1]
2. L¹[0, 1] ⊆ L²[0, 1]
3. f(x) = 1/x belongs to L²(1, ∞) but not to L¹(1, ∞)
4. L^∞[0, 1] is separable
Show answer
Answer: A — L²[0, 1] ⊆ L¹[0, 1]; C — f(x) = 1/x belongs to L²(1, ∞) but not to L¹(1, ∞)
(1) By Cauchy–Schwarz ‖f‖₁ ≤ ‖1‖₂‖f‖₂ = ‖f‖₂ on a set of measure 1. (2) x^−1/2 is in L¹(0, 1) (integral 2) but ∫x⁻¹ = ∞. (3) ∫₁^∞ x⁻² = 1 while ∫₁^∞ x⁻¹ = ∞. (4) The indicators 1_[0,t], t ∈ [0, 1], are uncountably many at mutual distance 1.
Let S = {p ≥ 1 : x^−1/3 ∈ Lᵖ(0, 1)}. The supremum of S is ____.
Numerical answer — type the value.
Show answer
Answer: 3
∫₀¹ x^−p/3 dx is finite iff p/3 < 1, i.e. p < 3. So S = [1, 3) and sup S = 3, not attained: at p = 3 the integral is ∫₀¹ dx/x = ∞.
Which statements about measurable functions on ℝ (Lebesgue measure) are true?
1. the pointwise limit of a sequence of measurable functions is measurable
2. every continuous function f: ℝ → ℝ is measurable
3. if f is measurable and g = f almost everywhere, then g is measurable
4. every subset of ℝ is Lebesgue measurable
Show answer
Answer: A — the pointwise limit of a sequence of measurable functions is measurable; B — every continuous function f: ℝ → ℝ is measurable; C — if f is measurable and g = f almost everywhere, then g is measurable
(1) lim fₙ = limsup fₙ = inf_k sup_n≥k fₙ, built from countable sups and infs. (2) {f > a} is open. (3) {g > a} differs from {f > a} by a subset of a null set, and every subset of a null set is measurable because Lebesgue measure is complete. (4) is false: Vitali’s set.
For 1 < p < ∞ and q with 1/p + 1/q = 1, which statement about Lᵖ[0, 1] is correct?
1. its dual space is isometrically isomorphic to L^q[0, 1]
2. it is not complete
3. its dual is L^p[0, 1] for every p
4. the continuous functions are not dense in it
Show answer
Answer: A — its dual space is isometrically isomorphic to L^{q}[0, 1]
The Riesz representation for Lᵖ: every bounded linear functional has the form f ↦ ∫fg for a unique g ∈ L^q, with norm ‖g‖_q. Lᵖ is complete (Riesz–Fischer) and C[0, 1] is dense for p < ∞. The dual is Lᵖ itself only for p = 2.