Comment. Countably additive measures
Comment. This quiz is designed to test your knowledge of countably additive measures on a sigma-algebra and related concepts.
Comment. Unless otherwise indicated, (X, A, mu) is a measure space.
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Question. If E_n is a sequence of measurable sets in X such that mu(E_n) goes to zero, and f: X -> R is a measurable function, what are the most general conditions under which one can assume the integral of f on E_n also goes to zero?
Correct Answer. One needs f to be bounded.
Comment. This is sufficient, but is rather strong.
Answer. One needs f to be non-negative.
Correct Answer. One needs f to be absolutely integrable.
Answer. One needs the E_n to be nested and decreasing.
Comment. This is not enough, even with monotone convergence, unless f is already absolutely integrable.
Answer. One needs f to be a simple function.
Comment. This is sufficient, but is far too strong.
Answer. One needs f to be continuous.
Comment. This is insufficient, if f is unbounded.
Answer. One needs f to be locally integrable.
Comment. This is insufficient, if f is unbounded.
Question. If mu and nu are non-negative measures, what does it mean for mu and nu to be mutually singular?
Answer. Given any measurable set E, at most one of mu(E) and nu(E) is non-zero.
Comment. This can easily fail for most pairs of mutually singular measures.
Answer. Given any measurable set E, exactly one of mu(E) and nu(E) is non-zero.
Comment. This can easily fail for most pairs of mutually singular measures.
Answer. Every measurable set E can be decomposed as E = F u G where mu(F) = 0 and nu(G) = 0.
Comment. This is a consequence of mutual singularity, but does not quite imply it.
Answer. Given any measurable set E, at most one of mu(E) and nu(E) is finite.
Correct Answer. There exists disjoint measurable sets F and G such that mu(E intersect F) = nu(E intersect G) = 0 for all measurable E.
Answer. Every set which is null in mu is non-null in nu, and vice versa.
Answer. There does not exist a set which is null in nu and mu simultaneously.
Question. If mu is a signed measure, how does one define the unsigned measure |mu|?
Correct Answer. |mu| is equal to the positive variation of mu plus the negative variation of mu.
Answer. |mu| is equal to +mu if mu is positive and -mu if mu is negative.
Comment. Most signed measures are neither positive nor negative, but need to be decomposed into components.
Answer. |mu| is the measure which gives each set E a measure of |mu(E)|.
Partially Correct Answer. |mu| is the measure with density |dmu(x)/dm(x)| with respect to Lebesgue measure dm.
Comment. This is only true when mu is absolutely continuous with respect to Lebesgue measure.
Answer. |mu| is the positive variation of mu.
Answer. |mu| is the difference of the positive and negative variations of mu.
Answer. |mu| is infinity if mu is an infinite measure, and is equal to |mu(X)| when mu is a finite measure.
Comment. Measures are not numbers.
Question. Let f be an absolutely integrable complex-valued function with respect to a measure of mu. What is the most precise relationship between the integral of |f| and the integral of f?
Correct Answer. The integral of f has magnitude less than or equal to the integral of |f|.
Answer. The magnitude of the integral of f is equal to the integral of |f|.
Answer. The integral of f has magnitude greater than or equal to the integral of |f|.
Answer. The real and imaginary parts of the integral of f are less than the integral of |f|.
Comment. This is true but is not the most precise statement one can make.
Answer. The sum of squares of the real and imaginary parts of the integral of f add up to the square of the integral of |f|.
Answer. Nothing can be said unless f has a fixed sign or phase.
Answer. The integral of f is strictly smaller in magnitude than the integral of |f|.
Answer. The integral of f lies between the integral of |f| and the negative integral of |f|.
Comment. This does not make sense since the integral of f is complex.
Question. What does it mean for a signed measure mu to be supported on a measurable set E?
Correct Answer. mu(F) = 0 whenever F is disjoint from E and measurable.
Answer. mu(E) = 0.
Partially Correct Answer. mu(X\E) = 0.
Comment. This would only be correct if mu was unsigned.
Answer. E is the smallest set such that mu(X\E) = 0.
Answer. E is the largest set such that mu(E) is non-zero.
Answer. mu(E) is non-zero.
Answer. mu(F) is non-zero whenever F is contained in E and measurable.
Question. Let E be a Lebesgue measurable subset of the real line. Which of the following statements is true?
Correct Answer. If E is bounded, then it has finite Lebesgue measure.
Answer. If E has finite Lebesgue measure, then it is bounded.
Answer. If E is bounded, then it is finite.
Answer. If E is unbounded, then it has infinite Lebesgue measure.
Answer. If E has zero measure, then it is bounded.
Answer. If E is uncountable, it has non-zero measure.
Answer. If E is uncountable, then it is unbounded.
Question. The support of a finite measure mu is
Answer. The largest set E which does not contain null sets.
Answer. The smallest set E whose complement is null.
Answer. The smallest set E whose complement is totally null.
Correct Answer. Not unique; a measure can have more than one support.
Answer. The union of all the sets of positive measure.
Answer. The intersection of all the sets of positive measure.
Answer. The intersection of all the sets of full measure.
Question. Let mu be an unsigned measure, and nu be a signed measure. Which of the following statements is true?
Correct Answer. If -mu <= nu <= mu, then nu is absolutely continuous with respect to mu.
Answer. If -mu <= nu <= mu, then mu is absolutely continuous with respect to nu.
Answer. If nu <= mu, then nu is absolutely continuous with respect to mu.
Answer. If nu is absolutely continuous with respect to mu. then nu <= mu.
Answer. If nu is absolutely continuous with respect to mu, then mu <= nu.
Answer. If 0 <= nu <= mu, then mu is absolutely continuous with respect to mu.
Answer. If nu is absolutely continuous with respect to mu, then we have -C mu <= nu <= C mu for some constant C.
Question. Let mu be an unsigned measure, and nu be a signed measure. What does it mean for nu to be absolutely continuous with respect to nu?
Answer. Every point {x} has a nu-measure of zero.
Comment. This is what it means for nu to be continuous, not absolutely continuous.
Answer. The Radon-Nikodym derivative d nu / d mu is a continuous function.
Answer. The Radon-Nikodym derivative d nu / d mu is a continuous function almost everywhere.
Correct Answer. Every set which has a mu-measure of zero also has a nu-measure of zero.
Answer. Every set which has a nu-measure of zero also has a mu-measure of zero.
Answer. For any measurable set E, nu(E) <= mu(E).
Answer. nu is supported on a set of positive mu-measure.
Answer. nu is not absolutely singular with respect to mu.