Lebesgue Measure and Integration Questions and Answers

Answer: A

The answer is \(m(M) = m(N) = m(O)\).

A countable set has Lebesgue measure 0, so L and every subset of it are null. N differs from M only by \(L \backslash M\), a null set, so \(m(N) = m(M)\).

O is the symmetric difference of L and M. It differs from M by removing \(L \cap M\) and adding \(L \backslash M\), both null, so \(m(O) = m(M)\) as well.

Adding or removing countably many points never changes Lebesgue measure, so no strict inequality is possible, which rules out the other three options.

Answer: C

\(\lambda\) does not define a signed measure.

Take \(A = \mathbf{R}\). The positive part of \(\sin x\) has infinite integral over \(\mathbf{R}\), and so does the negative part, so \(\int_{\mathbf{R}} \sin x\,d\mu\) is of the form \(\infty - \infty\) and is undefined.

A signed measure must be defined on every measurable set and may take at most one of the values \(+\infty\) and \(-\infty\). Here \(\lambda\) would have to be \(+\infty\) on \(\{\sin x \gt 0\}\) and \(-\infty\) on \(\{\sin x \lt 0\}\), so it is not a signed measure, and certainly not a measure, since it is negative on sets like \((\pi, 2\pi)\).

Answer: C

\(\nu_{1}\) and \(\nu_{2}\) are mutually singular.

\(\nu_{1}\) lives on \((1, \infty)\): it gives measure 0 to every subset of \((-\infty, 1]\). \(\nu_{2}\) lives on \([0, 1]\): it gives measure 0 to every subset of \((1, \infty)\). So the disjoint sets \((1, \infty)\) and \((-\infty, 1]\) separate them, which is the definition of mutual singularity.

Absolute continuity fails both ways: \(\nu_{2}((1, 2)) = 0\) but \(\nu_{1}((1, 2)) \gt 0\), and \(\nu_{1}([0, 1]) = 0\) but \(\nu_{2}([0, 1]) = e^{-1} \gt 0\). And Lebesgue measure gives positive mass to \((-2, -1)\), where both vanish, so it is not absolutely continuous with respect to either.

Answer: C

(P) \(\Leftrightarrow\) (Q).

For a measurable function, Lebesgue integrability is defined as \(\int f^{+} \lt \infty\) and \(\int f^{-} \lt \infty\). Since \(|f| = f^{+} + f^{-}\), this is the same as \(\int |f| \lt \infty\). So \(f\) is Lebesgue integrable if and only if \(|f|\) is.

This is a key difference from improper Riemann integration, where \(\sin x/x\) on \((0, \infty)\) converges conditionally while \(|\sin x/x|\) does not. Measurability is essential: for a non-measurable \(f\) with \(|f|\) constant, (Q) could hold without (P), but the question assumes \(f\) measurable.

Answer: B

The answer is for every \(\varepsilon \gt 0\) there is an open set U with \(E \subset U\) and \(m(U) \lt \varepsilon\).

Lebesgue measure is inner regular: for a measurable set, \(m(E) = \sup\{m(K) : K \subset E,\ K\) compact\(\}\). Since every compact subset has measure 0, \(m(E) = 0\). Outer regularity then gives, for each \(\varepsilon\), an open \(U \supset E\) with \(m(U) \lt m(E) + \varepsilon = \varepsilon\).

The other options do not follow. A null set need not be bounded (the integers, or the rationals). And not every null set is a \(G_{\delta}\) or an \(F_{\sigma}\): there are null subsets of the Cantor set that are not Borel at all.

Answer: B

\(\nu\) is a signed measure.

\(f\) is integrable (bounded, supported on \([0, 2\pi]\)), so \(\nu(A) = \int_{A} f\,d\mu\) is countably additive by the dominated convergence theorem and \(\nu(\emptyset) = 0\). It takes both signs: \(\nu([0, \pi]) = 2\) and \(\nu([\pi, 2\pi]) = -2\).

A measure must be non-negative, so \(\nu\) is not a measure (and so not 'both'). It is a finite signed measure, with Jordan decomposition given by the positive and negative parts of \(\sin x\).

Answer: D

The set with non-zero measure is the set of limit points of \(\mathbf{Q}\) in \(\mathbf{R}\).

Every real number is a limit of rationals, so the set of limit points of \(\mathbf{Q}\) is all of \(\mathbf{R}\), which has infinite Lebesgue measure.

The others have measure 0. \(\{n + 1/n\}\) and \(\{a + b\sqrt{2} : a, b \in \mathbf{Q}\}\) are countable, and countable sets are null. The Cantor set is uncountable, but it is left after removing intervals of total length \(\sum 2^{n-1}/3^{n} = 1\) from [0, 1], so its measure is 0.

Answer: C

\(\nu(\frac{1}{2}, \frac{3}{2}) = \) 0.

Split the interval at 1, where \(f\) changes sign:

\[ \begin{aligned} \nu\left(\tfrac{1}{2}, \tfrac{3}{2}\right) &= \int_{1/2}^{1} 1\, dx + \int_{1}^{3/2} (-1)\, dx \\ &= \tfrac{1}{2} - \tfrac{1}{2} = 0 \end{aligned} \]

The positive and negative parts cancel. The Lebesgue measure of \((0, 1)\) is 1, so A and B are the same wrong value, which would ignore the negative part, and \(-1/2\) counts only the negative part.

Answer: D

The false statement is that \(\lambda\) is singular with respect to \(\mu\) on M(E).

\(\lambda(A) = \int_{A \cap E} f\, d\mu\) is a measure on M (countable additivity follows from the monotone convergence theorem), and its restriction to \(M(E)\) is a measure too, so A and B hold. If \(\mu(A) = 0\) then \(\lambda(A) = 0\), so \(\lambda \ll \mu\) (C); \(f\) is its Radon-Nikodym derivative.

A measure that is absolutely continuous with respect to \(\mu\) can be singular with respect to it only if it is the zero measure. For a non-negative \(f\) that is not zero almost everywhere on E, \(\lambda\) is not singular, so D is the false statement.

Answer: D

\(f\) is absolutely continuous if \(f'\) exists and is continuous.

If \(f'\) is continuous on \([0, 1]\), it is bounded, say \(|f'| \le M\), and by the mean value theorem \(|f(x) - f(y)| \le M|x - y|\). A Lipschitz function is absolutely continuous: given \(\varepsilon\), take \(\delta = \varepsilon/M\). Equivalently, \(f(x) = f(0) + \int_{0}^{x} f'\), the defining property.

The other conditions are not enough. The Cantor function is continuous, hence uniformly continuous on \([0, 1]\), of bounded variation and Riemann integrable, yet it is not absolutely continuous, because it rises from 0 to 1 while its derivative is 0 almost everywhere.