Functions that respect measurable sets

Mathematics · Calculus & Analysis · ages 22-23

Name ______________________ Date ____________

A staircase rounds f down to the nearest multiple of 0.5, with no cap. At a point where f(x) = 1.8, what value does the staircase take?
Answer: ______________
Let f(x) = x squared on [-2, 2]. Which set is the preimage of the open ray of values above 1?
- The union of [-2, -1) and (1, 2]
- Only the right piece, the interval (1, 2]
- The middle interval (-1, 1) where x squared is small
- The closed interval [-1, 1] including both endpoints
To test whether a function f is measurable, what do you check?
- That f is continuous at every point of its domain
- That f takes only finitely many values on its domain
- That preimages of open rays are measurable sets
Measurable functions f_n converge at every point to a real-valued function f. Then f is always measurable.
Circle one: True False
Why do the powers x to the n on [0, 1] matter here?
- They show integrals always swap with limits
- They show pointwise limits can break continuity
- They show measurable functions stay continuous
The indicator of a measurable set is a measurable function. Is this correct?
Circle one: True False
Which rule builds simple functions that increase to a non-negative f?
- Use the constant 0 at every stage of the sequence
- Cap f at n and round down to multiples of 1 over 2 to the n
- Cap f at n and round down to multiples of 1 over n each time
- Round f up to the next multiple of 1 over 2 to the n
The set where f exceeds 3 is not measurable. What follows about f?
- Nothing, since single levels never matter
- That f is not measurable
- That f must be unbounded on its domain
A student claims the limit of measurable functions might fail the preimage test. What is the flaw?
- The limit rewrites with countable operations, which measurable sets survive
- Every limit of functions is automatically continuous
- Preimages are only defined for simple functions
f is the identity function on the interval from 0 to 4. What is the measure of the set where f exceeds 1?
Answer: ______________

Answer key

For grown-ups. Fold this page away before handing over the rest.

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1.5 · The multiples of 0.5 at or below 1.8 are 0, 0.5, 1 and 1.5. The largest is 1.5, so the staircase takes 1.5 there, just below f.
The union of [-2, -1) and (1, 2] · x squared exceeds 1 exactly when x is below -1 or above 1. Inside [-2, 2] that gives [-2, -1) together with (1, 2], a union of intervals, so the preimage is measurable.
That preimages of open rays are measurable sets · Measurability is a statement about preimages, and rays are enough to check.
True · Because the limit exists at every point, f can be rewritten with countable unions and intersections of preimages, which keeps it inside the class.
They show pointwise limits can break continuity · Continuous powers converge pointwise to a jump at 1, so continuity does not survive.
True · Each preimage is the empty set, the whole space, the set or its complement, and all of these are measurable.
Cap f at n and round down to multiples of 1 over 2 to the n · Capping at n and rounding down to multiples of 1 over 2 to the n keeps each stage below f. Halving the step keeps each stage at or above the last, and the stages rise to f. Steps of 1 over n are not nested, so that rule can drop.
That f is not measurable · One bad ray preimage already breaks the definition.
The limit rewrites with countable operations, which measurable sets survive · When the limit exists, it equals a supremum of infima of tails. Each infimum is a countable intersection of preimages and the supremum is a countable union, which measurable sets survive.
3 · Outputs above 1 come from the interval (1, 4], which has length 3.