Dear friends, I read in Kolmogorov-Fomin's that the following property of measurable real or complex valued functions ##\varphi,f## defined on measure space ##X##, proven in the text for ##\mu(X)<\infty## only, is also valid if ##X=\bigcup_n X_n## is not of finite measure, but it is the union of a countable sequence of measurable sets of finite measure ##X_n## (which we can suppose such that ##X_1\subset X_2\subset ...##): if ##\varphi## is Lebesgue integrable on ##X## and ##\forall x\in X\quad|f(x)|\leq\varphi(x)## then ##f## is Lebesgue integrable on ##X##.(adsbygoogle = window.adsbygoogle || []).push({});

Given the http://librarum.org/book/10022/159 [Broken] of Lebesgue integral ##\int_X g(x)d\mu:=\lim_n \int_{X_n}g(x)d\mu## for such a measure space, I know, from the property above for ##X_n## such that ##\mu(X_n)<\infty##, that if ##\int_{X_n}\varphi(x)d\mu## exists then ##\int_{X_n}f(x)d\mu## also exists, but how can we know that if ##\lim_n\int_{X_n}\varphi(x)d\mu## exists then ##\lim_n\int_{X_n}f(x)d\mu## exists?

##\infty## thanks!

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# Two conditions of existence for Lebesgue integral

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