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Quiz: Modes of convergence
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last edited
by Terence Tao 8 years, 11 months ago
This quiz is designed to test your knowledge of various types of convergence of sequences of functions. In all questions, the are realvalued, absolutely integrable functions on a measure space .
Note that some questions have multiple correct answers.
Discuss this quiz
(Key: correct, incorrect, partially correct.)
 If converges uniformly to , will imply converges pointwise to ?
 Yes, always.
 Yes if the domain has finite measure, but not in general.
 Yes if one passes to a subsequence of , but not in general.
 Yes if the functions are dominated by an absolutely integrable function, but not in general.
 Yes if the functions are nonnegative and monotone increasing, but not in general.
 Yes if the functions are equicontinuous, but not in general.
 None of the above.
 If converges pointwise to , does this imply converges uniformly to ?
 Yes, always.
 Yes if the domain has finite measure, but not in general.
 INCORRECT. A counterexample is the "shrinking bump".
 Yes if one passes to a subsequence of , but not in general.
 INCORRECT. A counterexample is the "sliding bump".
 Yes if the functions are dominated by an absolutely integrable function, but not in general.
 INCORRECT. A counterexample is the "shrinking bump".
 Yes if the functions are nonnegative and monotone increasing, but not in general.
 INCORRECT. A counterexample is the "widening bump".
 Yes if the functions are equicontinuous, but not in general.
 INCORRECT. This is not quite enough, consider for instance the "sliding bump". However, if the functions are uniformly equicontinuous and the domain is compact, then pointwise convergence implies uniform convergence.
 None of the above.
 If converges uniformly to , does this imply converges in the sense?
 Yes, always.
 INCORRECT. A counterexample is the "widening, flattening bump".
 Yes if the domain has finite measure, but not in general.
 CORRECT. This is one of two correct answers.
 Yes if one passes to a subsequence of , but not in general.
 INCORRECT. A counterexample is the "widening, flattening bump".
 Yes if the functions are dominated by an absolutely integrable function, but not in general.
 CORRECT. This is one of two correct answers.
 Yes if the functions are nonnegative and monotone increasing, but not in general.
 INCORRECT. Note that this does not even imply that is absolutely integrable.
 Yes if the functions are equicontinuous, but not in general.
 INCORRECT. A counterexample is the "widening, flattening bump".
 None of the above.
 If converges to f in the sense, does this imply converges to uniformly?
 Yes, always.
 INCORRECT. A counterexample is the "narrowing, increasingly tall bump".
 Yes if the domain has finite measure, but not in general.
 INCORRECT. A counterexample is the "narrowing, increasingly tall bump".
 Yes if one passes to a subsequence of f_n, but not in general.
 INCORRECT. A counterexample is the "narrowing, increasingly tall bump".
 Yes if the functions are dominated by an absolutely integrable function, but not in general.
 INCORRECT. A counterexample is the "narrowing, increasingly tall bump".
 Yes if the functions are nonnegative and monotone increasing, but not in general.
 INCORRECT. A counterexample is the "slowly widening bump".
 Yes if the functions are equicontinuous, but not in general.
 INCORRECT. A counterexample is the "sliding, narrowing, increasingly tall bump". However if the domain is compact and the functions are uniformly continuous, then we obtain uniform convergence.
 None of the above.
 If converges to in the sense, does this imply converges to pointwise a.e.?
 Yes, always.
 INCORRECT. A counterexample is the "typewriter sequence".
 Yes if the domain has finite measure, but not in general.
 INCORRECT. A counterexample is the "typewriter sequence".
 Yes if one passes to a subsequence of , but not in general.
 CORRECT. This is a consequence of the Chebyshev (or Markov) inequality.
 Yes if the functions are dominated by an absolutely integrable function, but not in general.
 INCORRECT. A counterexample is the "typewriter sequence".
 Yes if the functions are nonnegative and monotone increasing, but not in general.
 CORRECT. This is a consequence of the Chebyshev (or Markov) inequality.
 Yes if the functions are equicontinuous, but not in general.
 INCORRECT. This is not enough, for instance consider a discrete space with atoms of arbitrarily small measure. However if the domain is compact and the functions are uniformly continuous, then we obtain uniform convergence.
 None of the above.
 If converges to pointwise a.e., does this imply converges to in the sense?
 Yes, always.
 INCORRECT. A counterexample is the "sliding bump".
 Yes if the domain has finite measure, but not in general.
 INCORRECT. A counterexample is the "narrowing, increasingly tall bump".
 Yes if one passes to a subsequence of , but not in general.
 INCORRECT. A counterexample is the "sliding bump".
 Yes if the functions are dominated by an absolutely integrable function, but not in general.
 CORRECT. This is the Lebesgue dominated convergence theorem.
 Yes if the functions are nonnegative and monotone increasing, but not in general.
 PARTIALLY. The monotone convergence theorem implies that the integral of converges to the integral of , which (if is absolutely integrable) implies convergence.
 Yes if the functions are equicontinuous, but not in general.
 INCORRECT. A counterexample is the "sliding bump". However if the domain is compact and the functions are uniformly continuous, then we obtain uniform convergence.
 None of the above.
 If converges to pointwise a.e., does this imply converges to in measure?
 Yes, always.
 INCORRECT. A counterexample is the "sliding bump".
 Yes if the domain has finite measure, but not in general.
 CORRECT. This is a consequence of Egoroff's theorem.
 Yes if one passes to a subsequence of , but not in general.
 INCORRECT. A counterexample is the "sliding bump".
 Yes if the functions are dominated by an absolutely integrable function, but not in general.
 CORRECT. This is a consequence of the Lebesgue dominated convergence theorem.
 Yes if the functions are nonnegative and monotone increasing, but not in general.
 CORRECT. This is a consequence of continuity from below of the underlying measure.
 Yes if the functions are equicontinuous, but not in general.
 INCORRECT. A counterexample is the "sliding bump". However if the domain is compact and the functions are uniformly continuous, then we obtain uniform convergence.
 None of the above.
 If converges to in measure, does this imply converges to pointwise a.e.?
 Yes, always.
 INCORRECT. A counterexample is the "typewriter sequence".
 Yes if the domain has finite measure, but not in general.
 CORRECT. A counterexample is the "typewriter sequence".
 Yes if one passes to a subsequence of but not in general.
 INCORRECT. This is a consequence of countable subadditivity of the underlying measure.
 Yes if the functions are dominated by an absolutely integrable function, but not in general.
 INCORRECT. A counterexample is the "typewriter sequence".
 Yes if the functions are nonnegative and monotone increasing, but not in general.
 CORRECT. This is a consequence of monotonicity of the underlying measure.
 Yes if the functions are equicontinuous, but not in general.
 PARTIALLY. This is true under some mild assumptions, for instance that every nonempty open set has positive measure.
 None of the above.
 If converges to uniformly, does this imply converges to in measure?
 Yes, always.
 Yes if the domain has finite measure, but not in general.
 Yes if one passes to a subsequence of but not in general.
 Yes if the functions are dominated by an absolutely integrable function, but not in general.
 Yes if the functions are nonnegative and monotone increasing, but not in general.
 Yes if the functions are equicontinuous, but not in general.
 None of the above.
 If converges to in measure, does this imply converges to in uniformly?
 Yes, always.
 INCORRECT. A counterexample is the "narrowing bump".
 Yes if the domain has finite measure, but not in general.
 INCORRECT. A counterexample is the "narrowing bump".
 Yes if one passes to a subsequence of , but not in general.
 INCORRECT. A counterexample is the "narrowing bump".
 Yes if the functions are dominated by an absolutely integrable function, but not in general.
 INCORRECT. A counterexample is the "narrowing bump".
 Yes if the functions are nonnegative and monotone increasing, but not in general.
 INCORRECT. A counterexample is the "slowly widening bump".
 Yes if the functions are equicontinuous, but not in general.
 INCORRECT. A counterexample can be found if the space is discrete and has atoms of arbitrarily small measure. However if the domain is compact and the functions are uniformly continuous, then we obtain uniform convergence.
 None of the above.
 If converges to in the sense, does this imply converges to in measure?
 Yes, always.
 CORRECT. This is a consequence of the Chebyshev or Markov inequality.
 Yes if the domain has finite measure, but not in general.
 Yes if one passes to a subsequence of , but not in general.
 Yes if the functions are dominated by an absolutely integrable function, but not in general.
 Yes if the functions are nonnegative and monotone increasing, but not in general.
 Yes if the functions are equicontinuous, but not in general.
 None of the above.
 If converges to in measure, does this imply converges to in the sense?
 Yes, always.
 INCORRECT. A counterexample is given by the "narrowing, increasingly tall bump".
 Yes if the domain has finite measure, but not in general.
 INCORRECT. A counterexample is given by the "narrowing, increasingly tall bump".
 Yes if one passes to a subsequence of f_n, but not in general.
 INCORRECT. A counterexample is given by the "narrowing, increasingly tall bump".
 Yes if the functions are dominated by an absolutely integrable function, but not in general.
 CORRECT. This can be obtained (with some effort) from the Lebesgue dominated convergence theorem.
 Yes if the functions are nonnegative and monotone increasing, but not in general.
 INCORRECT. A counterexample is given by .
 Yes if the functions are equicontinuous, but not in general.
 INCORRECT. A counterexample can be found if the space is discrete and has atoms of arbitrarily small measure. However if the domain is compact and the functions are uniformly continuous, then we obtain convergence.
 None of the above.
 If converges to uniformly, does this imply converges to almost uniformly?
 Yes, always.
 Yes if the domain has finite measure, but not in general.
 Yes if one passes to a subsequence of , but not in general.
 Yes if the functions are dominated by an absolutely integrable function, but not in general.
 Yes if the functions are nonnegative and monotone increasing, but not in general.
 Yes if the functions are equicontinuous, but not in general.
 None of the above.
 If converges to almost uniformly, does this imply converges to uniformly?
 Yes, always.
 INCORRECT. A counterexample is the "narrowing bump".
 Yes if the domain has finite measure, but not in general.
 INCORRECT. A counterexample is the "narrowing bump".
 Yes if one passes to a subsequence of , but not in general.
 INCORRECT. A counterexample is the "narrowing bump".
 Yes if the functions are dominated by an absolutely integrable function, but not in general.
 INCORRECT. A counterexample is the "narrowing bump".
 Yes if the functions are nonnegative and monotone increasing, but not in general.
 INCORRECT. A counterexample is the "slowly widening bump".
 Yes if the functions are equicontinuous, but not in general.
 INCORRECT. A counterexample can be found if the space is discrete and has atoms of arbitrarily small measure. However if the domain is compact and the functions are uniformly continuous, then we obtain uniform convergence.
 None of the above.
 If converges to almost uniformly, does this imply converges to in the sense?
 Yes, always.
 INCORRECT. A counterexample is the "widening, flattening bump".
 Yes if the domain has finite measure, but not in general.
 INCORRECT. A counterexample is the "narrowing, increasingly tall bump".
 Yes if one passes to a subsequence of , but not in general.
 INCORRECT. A counterexample is the "widening, flattening bump".
 Yes if the functions are dominated by an absolutely integrable function, but not in general.
 CORRECT. This is a consequence of the dominated convergence theorem.
 Yes if the functions are nonnegative and monotone increasing, but not in general.
 INCORRECT. A counterexample is .
 Yes if the functions are equicontinuous, but not in general.
 INCORRECT. A counterexample can be found if the space is discrete and has atoms of arbitrarily small measure. However if the domain is compact and the functions are uniformly continuous, then we obtain uniform convergence.
 None of the above.
 If converges to in the sense, does this imply converges to almost uniformly?
 Yes, always.
 INCORRECT. A counterexample is the "typewriter sequence".
 Yes if the domain has finite measure, but not in general.
 INCORRECT. A counterexample is the "typewriter sequence".
 Yes if one passes to a subsequence of , but not in general.
 CORRECT. This is a consequence of the Chebyshev or Markov inequality.
 Yes if the functions are dominated by an absolutely integrable function, but not in general.
 INCORRECT. A counterexample is the "typewriter sequence".
 Yes if the functions are nonnegative and monotone increasing, but not in general.
 INCORRECT. This is a consequence of the Chebyshev or Markov inequality.
 Yes if the functions are equicontinuous, but not in general.
 INCORRECT. A counterexample can be provided by considering a slowly moving typewriter sequence. However if the domain is compact and the functions are uniformly continuous, then we obtain uniform convergence.
 None of the above.
 If converges to almost uniformly, does this imply converges to in measure?
 Yes, always.
 Yes if the domain has finite measure, but not in general.
 Yes if one passes to a subsequence of , but not in general.
 Yes if the functions are dominated by an absolutely integrable
 Yes if the functions are nonnegative and monotone increasing, but not in general.
 Yes if the functions are equicontinuous, but not in general.
 None of the above.
 If converges to in measure, does this imply converges to almost uniformly?
 Yes, always.
 INCORRECT. A counterexample is the "typewriter sequence".
 Yes if the domain has finite measure, but not in general.
 INCORRECT. A counterexample is the "typewriter sequence".
 Yes if one passes to a subsequence of , but not in general.
 CORRECT. This is a consequence of countable subadditivity of the underlying measure.
 Yes if the functions are dominated by an absolutely integrable function, but not in general.
 INCORRECT. A counterexample is the "typewriter sequence".
 Yes if the functions are nonnegative and monotone increasing, but not in general.
 INCORRECT. This is a consequence of monotonicity of the underlying measure.
 Yes if the functions are equicontinuous, but not in general.
 INCORRECT. A counterexample can be provided by considering a slowly moving typewriter sequence. However if the domain is compact and the functions are uniformly continuous, then we obtain uniform convergence.
 None of the above.
 If converges to pointwise a.e., does this imply converges to almost uniformly?
 Yes, always.
 INCORRECT. A counterexample is the "sliding bump".
 Yes if the domain has finite measure, but not in general.
 CORRECT. This is Egoroff's theorem.
 Yes if one passes to a subsequence of , but not in general.
 INCORRECT. A counterexample is the "sliding bump".
 Yes if the functions are dominated by an absolutely integrable function, but not in general.
 CORRECT. This can eventually be deduced from Egoroff's theorem.
 Yes if the functions are nonnegative and monotone increasing, but not in general.
 INCORRECT. A counterexample is provided by the characteristic functions of .
 Yes if the functions are equicontinuous, but not in general.
 INCORRECT. A counterexample can be provided by .
 None of the above.
 If converges to almost uniformly, does this imply converges to pointwise a.e.?
 Yes, always.
 Yes if the domain has finite measure, but not in general.
 Yes if one passes to a subsequence of , but not in general.
 Yes if the functions are dominated by an absolutely integrable function, but not in general.
 Yes if the functions are nonnegative and monotone increasing, but not in general.
 Yes if the functions are equicontinuous, but not in general.
 None of the above.
Score:
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Quiz: Modes of convergence

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