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Past exam of the integration course of the University of Oxford / 2021 / 3 / b / ii / Solution 2026-10-25

By part i,

χ_{E_{n}} ∣ g ∣ \to 0

almost everywhere, while

0 \leq χ_{E_{n}} ∣ g ∣ \leq ∣ g ∣

and

g

is Lebesgue integrable. The dominated convergence theorem gives

\int_{E_{n}} ∣ g (x) ∣ d x ⟶ 0.

To deduce absolute continuity of the Lebesgue integral, suppose the stated uniform conclusion failed for some

ε_{0} > 0

. For each

n

one could choose a measurable set

A_{n}

with

m (A_{n}) < 2^{- n}

but

\int_{A_{n}} ∣ g ∣ \geq ε_{0}

. Applying the result just proved to this sequence contradicts the lower bound. Hence

\forall ε > 0 \exists δ > 0 : m (A) < δ ⟹ \int_{A} ∣ g ∣ < ε

for every measurable set

A

.

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