When limits may pass inside the integral

Mathematics · Calculus & Analysis · ages 22-24

Name ______________________ Date ____________

On [0, 1], f_n equals 2 on [0, 1 minus 1/n] and 0 elsewhere. The f_n climb to a limit. What is the limit of their integrals?
Answer: ______________
On [0, infinity), f_n equals e to the minus x on [0, n] and 0 beyond n. The f_n climb to e to the minus x. What do their integrals approach?
- 0
- 1
- e
- Infinity
Non-negative functions climb steadily up to f. What does monotone convergence conclude?
- The integrals climb to the integral of the limit
- The limit function must be continuous
- Every function in the climb is bounded
Dominated convergence needs one integrable function g that sits above every |f_n|, the absolute values of the whole sequence.
Circle one: True False
Height 1 bumps on [2n, 2n+1] slide right along [0, infinity) as n grows. Which integrable function dominates them all?
- The constant function 1, which sits above every bump
- The limit function 0, since the bumps tend to 0
- None with a finite integral exists
If no integrable dominating function exists, dominated convergence does not apply, and the limit of the integrals may differ from the integral of the limit. Is this correct?
Circle one: True False
Which sequence on [0, 1] makes Fatou's inequality strict?
- The indicator of [0, 1 minus 1/n]
- n times the indicator of (0, 1/n)
- The constant function 1/n
- The powers x to the n
In the monotone convergence argument, which of these is a real use of monotonicity?
- To show each function in the sequence is continuous everywhere
- To guarantee the integrals climb along with the functions
- To make sure every function in the sequence is bounded above
Height 1 bumps on [2n, 2n+1] slide right. Their integrals stay at 1, but their pointwise limit integrates to 0. What explains the gap?
- No climbing and no integrable ceiling, so neither theorem applies
- The bump functions are not measurable
- The bumps take negative values somewhere on the line
On [0, 1], f_n equals n on (0, 1/n) and 0 elsewhere. What is the gap between the lower limit of the integrals and the integral of the pointwise limit?
Answer: ______________

Answer key

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When limits may pass inside the integral W1-mt_7n6ZDkHeBe-s1

2 · Each integral is 2 times (1 minus 1/n). The functions increase, so monotone convergence applies, and the integrals rise to 2.
1 · The f_n are non-negative and increase to e to the minus x, so monotone convergence applies. Each integral is 1 minus e to the minus n, which rises to 1.
The integrals climb to the integral of the limit · Increasing functions let the limit pass inside the integral.
True · One integrable g above every |f_n| is what keeps mass from escaping, so the swap of limit and integral holds.
None with a finite integral exists · Any ceiling must be at least 1 on every bump, and the bumps together cover infinite length on [0, infinity), so no integrable ceiling exists.
True · The theorem needs a dominating function, and without one the swap can fail.
n times the indicator of (0, 1/n) · n times the indicator of (0, 1/n) has integral 1 for every n but pointwise limit 0, so 0 is strictly below 1. The other sequences have integrals that tend to the integral of their limits.
To guarantee the integrals climb along with the functions · Climbing functions give climbing integrals, so the integrals have a limit that stays at or below the integral of f. Continuity and boundedness play no part.
No climbing and no integrable ceiling, so neither theorem applies · The bumps slide instead of climbing, so monotone convergence does not apply. Any function above every bump has infinite integral, so dominated convergence does not apply either. The bumps are measurable and never negative.
1 · Each integral is n times 1/n, which is 1, so the lower limit is 1. The pointwise limit is 0, with integral 0. The gap is 1.