Lebesgue Measure and Integration Questions and Answers

Q: The Lebesgue measure of the set \(A = \left\{0 \lt x \le 1 : x\sin\left(\frac{\pi}{2x}\right) \ge 0\right\}\) is

The measure is \(1 - \ln\sqrt{2}\).For \(x \gt 0\) the condition is \(\sin(\pi/2x) \ge 0\). Put \(t = \pi/2x\), which runs over \([\pi/2, \infty)\) as \(x\) runs over \((0, 1]\). The sine is non-negative on \([\pi/2, \pi]\) and on each \([2k\pi, (2k+1)\pi]\), which correspond to\[ \begin{aligned} x &\in [1/2, 1] \text{ (length } 1/2) \\ x &\in \left[\tfrac{1}{4k+2}, \tfrac{1}{4k}\right],\ k \ge 1 \end{aligned} \]Adding the lengths:\[ \begin{aligned} m(A) &= \tfrac{1}{2} + \sum_{k\ge 1}\left(\tfrac{1}{4k} - \tfrac{1}{4k+2}\right) \\ &= \tfrac{1}{2} + \tfrac{1}{2}\left(\tfrac{1}{2} - \tfrac{1}{3} + \tfrac{1}{4} - \cdots\right) \\ &= \tfrac{1}{2} + \tfrac{1}{2}(1 - \ln 2) = 1 - \ln\sqrt{2} \end{aligned} \]using \(1 - \frac{1}{2} + \frac{1}{3} - \cdots = \ln 2\). The value is about 0.653, so 0 and 1 are wrong, and \(\ln 2\) (about 0.693) is close to it but does not equal the sum of the lengths.

Q: Let \(\mathcal{N} \subseteq \mathbb{R}\) be a non-measurable set with respect to the Lebesgue measure on \(\mathbb{R}\). Consider the following statements:P: If \(M = \{x \in \mathcal{N} : x \text{ is irrational}\}\), then M is Lebesgue measurable.Q: The boundary of \(\mathcal{N}\) has positive Lebesgue outer measure.Then

P is FALSE and Q is TRUE.P: \(\mathcal{N}\) is the union of M and \(\mathcal{N} \cap \mathbb{Q}\), which is countable and so measurable with measure zero. If M were measurable, \(\mathcal{N}\) would be a union of two measurable sets and hence measurable, a contradiction. So M is not measurable.Q: write \(\mathcal{N}\) as the union of its interior and \(\mathcal{N} \cap \partial\mathcal{N}\). The interior is open and so measurable. If the boundary had outer measure zero, \(\mathcal{N} \cap \partial\mathcal{N}\) would be a subset of a null set, hence measurable because Lebesgue measure is complete, and \(\mathcal{N}\) would be measurable. So the boundary must have positive outer measure.

Q: For \(k \in \mathbb{N}\), let \(E_{k}\) be a measurable subset of [0,1] with Lebesgue measure \(\frac{1}{k^{2}}\). Define\[ \begin{aligned} E &= \bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}E_{k} \text{ and} \\ F &= \bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}E_{k} \end{aligned} \]Consider the following statements:P: Lebesgue measure of E is equal to zero.Q: Lebesgue measure of F is equal to zero.Then

Both P and Q are TRUE.E is the lim sup of the sets \(E_{k}\), the points lying in infinitely many of them. Since\[ \begin{aligned} \sum_{k=1}^{\infty} m(E_{k}) &= \sum_{k=1}^{\infty}\frac{1}{k^{2}} \\ &= \frac{\pi^{2}}{6} \lt \infty, \end{aligned} \]the Borel-Cantelli lemma gives m(E) = 0. Directly, \(E \subseteq \bigcup_{k \ge n}E_{k}\), whose measure is at most \(\sum_{k \ge n}1/k^{2} \to 0\).F is the lim inf, the points lying in all but finitely many \(E_{k}\). Every such point lies in infinitely many, so \(F \subseteq E\) and m(F) = 0 as well. The other options would need the series of measures to diverge.

Q: Let \(f : [0, 1] \to \mathbb{R}\) be a function. Which one of the following is a sufficient condition for \(f\) to be Lebesgue measurable?

A sufficient condition is \(f\) is continuous almost everywhere on [0, 1].If \(f\) is continuous outside a null set \(N\), then \(f\) restricted to \([0, 1] \setminus N\) is continuous, hence Borel measurable there. Every subset of \(N\) is Lebesgue measurable because Lebesgue measure is complete, so \(\{f \gt c\}\) is measurable for every \(c\).The others fail. Take a non-measurable set \(V\) and \(f = 1\) on \(V\), \(-1\) off it: then \(|f| = 1\) is measurable and \(-1 \le f \le 1\) is squeezed between continuous functions, but \(f\) is not measurable. And \(f(x) = x\) on \(V\), \(-x\) off it, takes each value at most twice, so every level set is finite and measurable, yet \(f\) is not measurable.

Q: Let C denote the Cantor set and \(f : [0,1] \to \mathbb{R}\) be defined as follows:\[ f(x) = \begin{cases} x^{2026} & \text{for } x \in C \\ \cos(\pi x) & \text{for } x \in [0, \frac{1}{2}] \setminus C \\ \sin(\pi x) & \text{for } x \in [\frac{1}{2}, 1] \setminus C \end{cases} \]The value of the Lebesgue integral of \(f(x)\) over the interval [0,1] is equal to

The Lebesgue integral equals \(\frac{2}{\pi}\).The Cantor set has Lebesgue measure zero, so the values of f on it, including \(x^{2026}\), do not affect the integral. Off the Cantor set f is \(\cos \pi x\) on \([0, \frac{1}{2}]\) and \(\sin \pi x\) on \([\frac{1}{2}, 1]\):\[ \begin{aligned} \int_0^{1/2} \cos \pi x\, dx &= \frac{1}{\pi} \\ \int_{1/2}^1 \sin \pi x\, dx &= \frac{1}{\pi} \end{aligned} \]The sum is \(\frac{2}{\pi}\). \(\frac{1}{\pi}\) counts only one of the two pieces, 0 would follow only if the whole interval were negligible, and \(\frac{3}{\pi}\) wrongly adds a contribution from the Cantor set.

Q: The Lebesgue integral of the function \(f : [0, 1] \to \mathbf{R}\) defined by :\[ f(x) = \begin{cases} 1/\sqrt[3]{x} & \text{if } 0 \lt x \le 1 \\ 0 & \text{if } x = 0 \end{cases} \]is :

The integral is 3/2.\(f(x) = x^{-1/3}\) is non-negative and the single point \(x = 0\) has measure zero, so the Lebesgue integral equals the improper Riemann integral:\[ \begin{aligned} \int_{0}^{1} x^{-1/3}\,dx &= \left[\frac{x^{2/3}}{2/3}\right]_{0}^{1} \\ &= \frac{3}{2} \end{aligned} \]The integral is finite because the exponent \(1/3\) is less than 1. The value 3/4 is \(\int_{0}^{1} x^{1/3}dx\), the wrong sign of exponent; 2/3 inverts the answer; and 4/3 does not come from either.

Q: Let \(\mu^{*}\) be an outer measure on \(H(\mathbf{R})\). Then \(E \in H(\mathbf{R})\) is \(\mu^{*}\)-measurable if :where \(H(\mathbf{R})\)-hereditary \(\sigma\)-ring.

E is \(\mu^{*}\)-measurable if for each \(A\), \(\mu^{*}(A) = \mu^{*}(A \cap E) + \mu^{*}(A \cap E^{C})\).This is Caratheodory's criterion: E must split every test set A additively. The family of such sets is a \(\sigma\)-ring on which \(\mu^{*}\) is a measure.The inequality \(\mu^{*}(A) \le \mu^{*}(A \cap E) + \mu^{*}(A \cap E^{C})\) always holds, by subadditivity of an outer measure, so options C and D say nothing about E. Requiring the equation for only some A (option B) is far too weak: \(A = \emptyset\) satisfies it for every E.

Q: Consider the following statements :I. If \(v_{1}\), \(v_{2}\) and \(\mu\) are measures and \(v_{1} \perp \mu\), \(v_{2} \perp \mu\), then \(v_{1} + v_{2} \perp \mu\).II. If \(v\) and \(\mu\) are measures such that \(v \ll \mu\) and \(v \perp \mu\), then \(v\) is identically zero.

Both I and II are true.I: \(v_{1} \perp \mu\) gives a set \(A_{1}\) with \(\mu(A_{1}) = 0\) carrying all of \(v_{1}\), that is \(v_{1}(A_{1}^{c}) = 0\); similarly \(A_{2}\) for \(v_{2}\). Then \(A = A_{1} \cup A_{2}\) has \(\mu(A) = 0\) and \((v_{1} + v_{2})(A^{c}) = 0\), so \(v_{1} + v_{2} \perp \mu\).II: \(v \perp \mu\) gives a set A with \(\mu(A) = 0\) and \(v(A^{c}) = 0\). Absolute continuity \(v \ll \mu\) then gives \(v(A) = 0\) too, so \(v(X) = v(A) + v(A^{c}) = 0\). A measure both absolutely continuous and singular with respect to \(\mu\) is zero; this is what makes the Lebesgue decomposition unique.

The Lebesgue measure of the set \(A = \left\{0 \lt x \le 1 : x\sin\left(\frac{\pi}{2x}\right) \ge 0\right\}\) is

A.0
B.1
C.\(\ln 2\)
D.\(1 - \ln\sqrt{2}\)

Exam Year(s)

GATE Mathematics - 2012,

Answer: D

The measure is \(1 - \ln\sqrt{2}\).

For \(x \gt 0\) the condition is \(\sin(\pi/2x) \ge 0\). Put \(t = \pi/2x\), which runs over \([\pi/2, \infty)\) as \(x\) runs over \((0, 1]\). The sine is non-negative on \([\pi/2, \pi]\) and on each \([2k\pi, (2k+1)\pi]\), which correspond to

\[ \begin{aligned} x &\in [1/2, 1] \text{ (length } 1/2) \\ x &\in \left[\tfrac{1}{4k+2}, \tfrac{1}{4k}\right],\ k \ge 1 \end{aligned} \]

Adding the lengths:

\[ \begin{aligned} m(A) &= \tfrac{1}{2} + \sum_{k\ge 1}\left(\tfrac{1}{4k} - \tfrac{1}{4k+2}\right) \\ &= \tfrac{1}{2} + \tfrac{1}{2}\left(\tfrac{1}{2} - \tfrac{1}{3} + \tfrac{1}{4} - \cdots\right) \\ &= \tfrac{1}{2} + \tfrac{1}{2}(1 - \ln 2) = 1 - \ln\sqrt{2} \end{aligned} \]

using \(1 - \frac{1}{2} + \frac{1}{3} - \cdots = \ln 2\). The value is about 0.653, so 0 and 1 are wrong, and \(\ln 2\) (about 0.693) is close to it but does not equal the sum of the lengths.

Discuss in Forum

Let \(\mathcal{N} \subseteq \mathbb{R}\) be a non-measurable set with respect to the Lebesgue measure on \(\mathbb{R}\). Consider the following statements:

P: If \(M = \{x \in \mathcal{N} : x \text{ is irrational}\}\), then M is Lebesgue measurable.
Q: The boundary of \(\mathcal{N}\) has positive Lebesgue outer measure.

Then

A.both P and Q are TRUE
B.P is FALSE and Q is TRUE
C.P is TRUE and Q is FALSE
D.both P and Q are FALSE

Exam Year(s)

GATE Mathematics - 2023,

Answer: B

P is FALSE and Q is TRUE.

P: \(\mathcal{N}\) is the union of M and \(\mathcal{N} \cap \mathbb{Q}\), which is countable and so measurable with measure zero. If M were measurable, \(\mathcal{N}\) would be a union of two measurable sets and hence measurable, a contradiction. So M is not measurable.

Q: write \(\mathcal{N}\) as the union of its interior and \(\mathcal{N} \cap \partial\mathcal{N}\). The interior is open and so measurable. If the boundary had outer measure zero, \(\mathcal{N} \cap \partial\mathcal{N}\) would be a subset of a null set, hence measurable because Lebesgue measure is complete, and \(\mathcal{N}\) would be measurable. So the boundary must have positive outer measure.

Discuss in Forum

For \(k \in \mathbb{N}\), let \(E_{k}\) be a measurable subset of [0,1] with Lebesgue measure \(\frac{1}{k^{2}}\). Define

\[ \begin{aligned} E &= \bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}E_{k} \text{ and} \\ F &= \bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}E_{k} \end{aligned} \]

Consider the following statements:

P: Lebesgue measure of E is equal to zero.
Q: Lebesgue measure of F is equal to zero.

Then

A.both P and Q are TRUE
B.both P and Q are FALSE
C.P is TRUE but Q is FALSE
D.Q is TRUE but P is FALSE

Exam Year(s)

GATE Mathematics - 2023,

Answer: A

Both P and Q are TRUE.

E is the lim sup of the sets \(E_{k}\), the points lying in infinitely many of them. Since

\[ \begin{aligned} \sum_{k=1}^{\infty} m(E_{k}) &= \sum_{k=1}^{\infty}\frac{1}{k^{2}} \\ &= \frac{\pi^{2}}{6} \lt \infty, \end{aligned} \]

the Borel-Cantelli lemma gives m(E) = 0. Directly, \(E \subseteq \bigcup_{k \ge n}E_{k}\), whose measure is at most \(\sum_{k \ge n}1/k^{2} \to 0\).

F is the lim inf, the points lying in all but finitely many \(E_{k}\). Every such point lies in infinitely many, so \(F \subseteq E\) and m(F) = 0 as well. The other options would need the series of measures to diverge.

Discuss in Forum

Let \(f : [0, 1] \to \mathbb{R}\) be a function. Which one of the following is a sufficient condition for \(f\) to be Lebesgue measurable?

A.\(|f|\) is a Lebesgue measurable function
B.There exist continuous functions \(g, h : [0, 1] \to \mathbb{R}\) such that \(g \le f \le h\) on [0, 1]
C.\(f\) is continuous almost everywhere on [0, 1]
D.For each \(c \in \mathbb{R}\), the set \(\{x \in [0, 1] : f(x) = c\}\) is Lebesgue measurable

Exam Year(s)

GATE Mathematics - 2024,

Answer: C

A sufficient condition is \(f\) is continuous almost everywhere on [0, 1].

If \(f\) is continuous outside a null set \(N\), then \(f\) restricted to \([0, 1] \setminus N\) is continuous, hence Borel measurable there. Every subset of \(N\) is Lebesgue measurable because Lebesgue measure is complete, so \(\{f \gt c\}\) is measurable for every \(c\).

The others fail. Take a non-measurable set \(V\) and \(f = 1\) on \(V\), \(-1\) off it: then \(|f| = 1\) is measurable and \(-1 \le f \le 1\) is squeezed between continuous functions, but \(f\) is not measurable. And \(f(x) = x\) on \(V\), \(-x\) off it, takes each value at most twice, so every level set is finite and measurable, yet \(f\) is not measurable.

Discuss in Forum

Let C denote the Cantor set and \(f : [0,1] \to \mathbb{R}\) be defined as follows:

\[ f(x) = \begin{cases} x^{2026} & \text{for } x \in C \\ \cos(\pi x) & \text{for } x \in [0, \frac{1}{2}] \setminus C \\ \sin(\pi x) & \text{for } x \in [\frac{1}{2}, 1] \setminus C \end{cases} \]

The value of the Lebesgue integral of \(f(x)\) over the interval [0,1] is equal to

A.\(\frac{2}{\pi}\)
B.\(\frac{1}{\pi}\)
C.\(\frac{3}{\pi}\)
D.0

Exam Year(s)

GATE Mathematics - 2026,

Answer: A

The Lebesgue integral equals \(\frac{2}{\pi}\).

The Cantor set has Lebesgue measure zero, so the values of f on it, including \(x^{2026}\), do not affect the integral. Off the Cantor set f is \(\cos \pi x\) on \([0, \frac{1}{2}]\) and \(\sin \pi x\) on \([\frac{1}{2}, 1]\):

\[ \begin{aligned} \int_0^{1/2} \cos \pi x\, dx &= \frac{1}{\pi} \\ \int_{1/2}^1 \sin \pi x\, dx &= \frac{1}{\pi} \end{aligned} \]

The sum is \(\frac{2}{\pi}\). \(\frac{1}{\pi}\) counts only one of the two pieces, 0 would follow only if the whole interval were negligible, and \(\frac{3}{\pi}\) wrongly adds a contribution from the Cantor set.

Discuss in Forum

Consider the functions \(f, g : [0, 1] \to \mathbf{R}\) defined by :

\[ f(x) = \begin{cases} 1 & \text{if } x \text{ is rational} \\ 0 & \text{if } x \text{ is irrational} \end{cases} \]

and

\[ g(x) = \begin{cases} 1/x & \text{if } 0 \lt x \le 1 \\ 0 & \text{if } x = 0 \end{cases} \]

then :

A.both \(f\) and \(g\) are not Lebesgue integrable
B.\(f\) is Riemann integrable and \(g\) is Lebesgue integrable
C.\(f\) is not Riemann integrable and \(g\) is Lebesgue integrable
D.\(f\) is Lebesgue integrable and \(g\) is not Lebesgue integrable

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

\(f\) is Lebesgue integrable and \(g\) is not Lebesgue integrable.

\(f\) is the Dirichlet function. It equals 0 except on the rationals, a countable set of measure zero, so it is Lebesgue integrable with integral 0. (It is not Riemann integrable, since every upper sum is 1 and every lower sum 0.)

For \(g\), \(\int_{\epsilon}^{1} dx/x = -\log\epsilon \to \infty\), so \(\int_{[0,1]} g = \infty\) and \(g\) is not Lebesgue integrable. That rules out the options calling \(g\) integrable, and option A, which also wrongly denies that \(f\) is integrable.

Discuss in Forum

The Lebesgue integral of the function \(f : [0, 1] \to \mathbf{R}\) defined by :

\[ f(x) = \begin{cases} 1/\sqrt[3]{x} & \text{if } 0 \lt x \le 1 \\ 0 & \text{if } x = 0 \end{cases} \]

is :

A.\(\dfrac{3}{4}\)
B.\(\dfrac{2}{3}\)
C.\(\dfrac{3}{2}\)
D.\(\dfrac{4}{3}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The integral is 3/2.

\(f(x) = x^{-1/3}\) is non-negative and the single point \(x = 0\) has measure zero, so the Lebesgue integral equals the improper Riemann integral:

\[ \begin{aligned} \int_{0}^{1} x^{-1/3}\,dx &= \left[\frac{x^{2/3}}{2/3}\right]_{0}^{1} \\ &= \frac{3}{2} \end{aligned} \]

The integral is finite because the exponent \(1/3\) is less than 1. The value 3/4 is \(\int_{0}^{1} x^{1/3}dx\), the wrong sign of exponent; 2/3 inverts the answer; and 4/3 does not come from either.

Discuss in Forum

Let \(\mu^{*}\) be an outer measure on \(H(\mathbf{R})\). Then \(E \in H(\mathbf{R})\) is \(\mu^{*}\)-measurable if :

where \(H(\mathbf{R})\)-hereditary \(\sigma\)-ring.

A.for each \(A \in H(\mathbf{R})\), \(\mu^{*}(A) = \mu^{*}(A \cap E) + \mu^{*}(A \cap E^{C})\)
B.for some \(A \in H(\mathbf{R})\), \(\mu^{*}(A) = \mu^{*}(A \cap E) + \mu^{*}(A \cap E^{C})\)
C.for each \(A \in H(\mathbf{R})\), \(\mu^{*}(A) \le \mu^{*}(A \cap E) + \mu^{*}(A \cap E^{C})\)
D.for some \(A \in H(R)\), \(\mu^{*}(A) \le \mu^{*}(A \cap E) + \mu^{*}(A \cap E^{C})\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

E is \(\mu^{*}\)-measurable if for each \(A\), \(\mu^{*}(A) = \mu^{*}(A \cap E) + \mu^{*}(A \cap E^{C})\).

This is Caratheodory's criterion: E must split every test set A additively. The family of such sets is a \(\sigma\)-ring on which \(\mu^{*}\) is a measure.

The inequality \(\mu^{*}(A) \le \mu^{*}(A \cap E) + \mu^{*}(A \cap E^{C})\) always holds, by subadditivity of an outer measure, so options C and D say nothing about E. Requiring the equation for only some A (option B) is far too weak: \(A = \emptyset\) satisfies it for every E.

Discuss in Forum

Let E and F be measurable sets, \(f \in L(E, \mu)\) and \(\mu(E \,\Delta\, F) = 0\), then :

A.\(f \in L(F, \mu)\) and \(\int_{E} f\, d\mu = \int_{F} f\, d\mu\)
B.\(f \in L(F, \mu)\) and \(\int_{E} f\, d\mu \lt \int_{F} f\, d\mu\)
C.\(f \in L(F, \mu)\) and \(\int_{E} f\, d\mu \gt \int_{F} f\, d\mu\)
D.\(f \notin L(F, \mu)\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

\(f \in L(F, \mu)\) and \(\int_{E} f\,d\mu = \int_{F} f\,d\mu\).

\(\mu(E \,\Delta\, F) = 0\) means E and F differ by a null set: \(E \setminus F\) and \(F \setminus E\) both have measure zero. Integrals ignore null sets, so

\[ \begin{aligned} \int_{F} f &= \int_{E \cap F} f + \int_{F \setminus E} f \\ &= \int_{E \cap F} f + 0 \end{aligned} \]

and similarly \(\int_{E} f = \int_{E \cap F} f\). Since \(|f|\) is integrable over E, it is integrable over \(E \cap F\) and hence over F. A strict inequality between the two integrals, or non-integrability over F, is therefore impossible.

Discuss in Forum

10.

Consider the following statements :

I. If \(v_{1}\), \(v_{2}\) and \(\mu\) are measures and \(v_{1} \perp \mu\), \(v_{2} \perp \mu\), then \(v_{1} + v_{2} \perp \mu\).
II. If \(v\) and \(\mu\) are measures such that \(v \ll \mu\) and \(v \perp \mu\), then \(v\) is identically zero.

A.Only I is true
B.Only II is true
C.Both I and II are not true
D.Both I and II are true

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

Both I and II are true.

I: \(v_{1} \perp \mu\) gives a set \(A_{1}\) with \(\mu(A_{1}) = 0\) carrying all of \(v_{1}\), that is \(v_{1}(A_{1}^{c}) = 0\); similarly \(A_{2}\) for \(v_{2}\). Then \(A = A_{1} \cup A_{2}\) has \(\mu(A) = 0\) and \((v_{1} + v_{2})(A^{c}) = 0\), so \(v_{1} + v_{2} \perp \mu\).

II: \(v \perp \mu\) gives a set A with \(\mu(A) = 0\) and \(v(A^{c}) = 0\). Absolute continuity \(v \ll \mu\) then gives \(v(A) = 0\) too, so \(v(X) = v(A) + v(A^{c}) = 0\). A measure both absolutely continuous and singular with respect to \(\mu\) is zero; this is what makes the Lebesgue decomposition unique.

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