Let ((X, mathscr{A}, mu)) be a measure space. Show the following variant of Theorem 9.6. If (u_{n}

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 Let \((X, \mathscr{A}, \mu)\) be a measure space. Show the following variant of Theorem 9.6. If \(u_{n} \geqslant 0\) are measurable functions such that for some \(u\) we have
\[
\exists K \in \mathbb{N} \quad \forall x: u_{n+K}(x) \uparrow u(x) \text { as } n ightarrow \infty
\]
then \(u \geqslant 0\) is measurable and \(\int u_{n} d \mu \uparrow \int u d \mu\).
Show that we cannot replace the above condition with
\[
\forall x \quad \exists K \in \mathbb{N}: u_{n+K}(x) \uparrow u(x) \text { as } n ightarrow \infty
\]

Data from theorem 9.6

(Beppo Levi) Let (X, A, u) be a measure space. For an increasing sequence of functions (Un)neN CM(A), 0un <Step 1. Claim: u, w M(A), u

where we use the definition of the sets B,, (for the penultimate estimate) and the fact that 1 BM, < 1 (in

Step 4. In the estimate which we established in Step 3, we can go to the supremum over all feE+(A) with f

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