The Lebesgue Integral and What It Repairs · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Slicing the range instead of the domain

Mathematics · Calculus & Analysis · ages 22-23
Name ______________________   Date ____________

  1. Indicator of the interval from 1 to 4 integrates to its length. What is it?

    Answer: ______________

  2. Simple function equals 2 on a set of measure 3. What is its integral?

    • 6
    • 5
    • 3
    • 2

  3. How do you integrate a simple function?

    • Take the largest value it reaches
    • Add value times measure over its pieces
    • Count how many pieces it has

  4. What is the Lebesgue integral over [0, 1] of the indicator of the rationals?

    • 0
    • 1
    • It has no integral

  5. A bounded function on [0, 1] is Riemann integrable. What follows for its Lebesgue integral?

    • It exists and equals the Riemann value
    • It exists but is always larger
    • It can never exist

  6. If Riemann integrable on a bounded interval, the Lebesgue integral exists and values agree. Is this correct?

    Circle one:   True   False

  7. Why does Riemann fail for the Dirichlet indicator?

    • Function is unbounded
    • Domain is empty
    • Upper sums stay 1 and lower stay 0 on every partition
    • Integrals always exist

  8. Riemann sums for the rationals indicator stall because every subinterval holds both rational and irrational points.

    Circle one:   True   False

  9. A student says the rationals indicator should integrate to 1 since rationals sit everywhere. What is the flaw?

    • Being everywhere is not the same as having positive measure
    • The rationals cannot be listed at all
    • Indicator functions can never be integrated

  10. Constant 4 on the interval from 0 to 2. Lebesgue integral is what?

    Answer: ______________

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Answer key

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

Slicing the range instead of the domain W1-mt_KO92nLc5YU-s1

  1. 3 · Length 4 minus 1 equals 3.
  2. 6 · 6 is correct since 2 times 3 equals 6, while 5, 3, and 2 miss the product.
  3. Add value times measure over its pieces · Each level contributes its height times the size of its piece.
  4. 0 · The rationals have measure zero, so only the zero level counts.
  5. It exists and equals the Riemann value · The new integral extends the old one without changing its values.
  6. True · Lebesgue extends Riemann on bounded intervals.
  7. Upper sums stay 1 and lower stay 0 on every partition · Upper sums stay 1 and lower stay 0 on every partition is correct since each interval holds both kinds of points, while boundedness holds and existence fails.
  8. True · Lower sums stick at 0 while upper sums stick at 1.
  9. Being everywhere is not the same as having positive measure · Density differs from size: the rationals are dense but have measure zero.
  10. 8 · 4 times length 2 equals 8.
Worksheet · LightMySky