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Let $X$ be a compact Hausdorff space and $\mu$ be a complex Borel measure on $X$ having finite total variation. Let $B(X)$ be the set of all complex valued bounded Borel measurable functions on $X.$ Let $f_{\alpha}, f \in B(X)$ be such that $f_{\alpha} \to f$ pointwise and $\sup\limits_{\alpha} \left \{\left |f_{\alpha} (x) \right |\ |\ x \in X \right \} \lt \infty.$ Then show that $\int f_{\alpha}\ d\mu \to \int f\ d\mu.$

We have $$\left | \int f_{\alpha}\ d\mu - \int f\ d\mu \right | = \left |\int (f_{\alpha} - f)\ d\mu \right | \leq \int |f_{\alpha} - f|\ d |\mu|.$$ Now how does the given conditions conclude the result?

EDIT $:$ If we have a sequence instead of a net then we are through by the virtue of DCT. But I don't have any idea whether DCT holds for pointwise convergence of a net of measurable functions which are uniformly bounded by an integrable function.

ACB
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1 Answers1

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Without further assumptions this cannot hold. A counterexample is given in the first answer here. If one additionally assumes that $(f_\alpha)_\alpha$ is monotone, then the conclusion is true, and a proof may be found here.

AlephBeth
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