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Yahir Tucker

Yahir Tucker

Answered question

2022-06-15

Let E ⊂ R be a Borel set and suppose that ( f k ) k ∈ N is a sequence of Lebesgue measurable functions defined on E. Assume that lim k → + ∞ f k ( x ) = f ( x ) for almost every x and that there exists a Lebesgue integrable function g such that | f k ( x ) | ≤ g ( x ) for almost every x ∈ E and every k ∈ N . Fix ε > 0 and define A k ε = { x ∈ E : | f k ( x ) − f ( x ) | ≥ ε }. Is it true that
lim j → + ∞ | ⋃ k ≥ j A k ε | = 0 ?
( | ⋅ | is the Lebesgue measure)
The only thing that I noticed is that | A k ε | → k → + ∞ 0, as a consequence of Lebesgue's dominated convergence theorem.
Does anyone know if this is true or not (and why?)?

Answer & Explanation

luisjoseblash2

luisjoseblash2

Beginner2022-06-16Added 16 answers

True. For every j ∈ { 1 , 2 , … } define B j := ⋃ k ≥ j A k ε . Then B 1 ⊇ B 2 ⊇ B 2 ⊇ ⋯, and λ ( ⋂ j = 1 ∞ B j ) = λ ( | f k ( x ) − f ( x ) | ≥ ε  i . o . ) = 0, since lim k → ∞ f k ( x ) = f ( x ) almost everywhere. So, if we can demonstrate that λ ( B 1 ) < ∞, then by the continuity of measures lim j → ∞ λ ( B j ) = 0. In fact, λ ( B 1 ) ≤ λ ( g ≥ ε 2 ) < ∞: the first inequality is due to the fact that if x ∈ B 1 , then | f k ( x ) − f ( x ) | ≥ ε for some k ∈ { 1 , 2 , … }, and almost everywhere | f k ( x ) | , | f ( x ) | ≤ g ( x ); Moreover, the fact that g is integrable accounts for the second inequality.

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