fix: Ex. 1.2.13 indicator limits need ae convergence - #582
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Pointwise convergence of indicators at every x does not force a measurable limit set ([0,1] \ Q counterexample). Use ae convergence via IsNull failure sets; (iii) assumes measurable E₀. Co-authored-by: Cursor <cursoragent@cursor.com>
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Exercise 1.2.13 in the text defines pointwise convergence of sets via
pointwise convergence of the indicator functions, and then asks:
(i) if the E_n are measurable and converge pointwise to E, show E is
measurable;
(ii) if additionally the E_n lie in a measurable F of finite measure,
show m(E_n) -> m(E);
(iii) give a counterexample when the E_n are not contained in a set of
finite measure.
Throughout, E is given and the convergence is pointwise everywhere, not
almost everywhere. teorth#582 replaced this with an a.e. formulation and, in
doing so, moved E from a hypothesis into the goal, making (i) and (ii)
false: for E_n alternating between the empty set and the whole space
(resp. F), the indicators oscillate at every point, so no measurable E₀
has 1_{E_n} -> 1_{E₀} off a null set.
This restores the previous statements, which match the text.
Co-Authored-By: Claude Opus 5 (1M context) <noreply@anthropic.com>
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Summary
xdoes not makeE₀Lebesgue measurable ([0,1] \ ℚcounterexample).IsNullfailure set; existential measurableE₀.E₀and ae convergence.Fixes part of #517.
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lake build(signature-only)Made with Cursor