# Two conditions of existence for Lebesgue integral

1. Oct 16, 2014

### DavideGenoa

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 $X$.

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!

Last edited by a moderator: May 7, 2017
2. Oct 17, 2014

### mathman

I haven't worked out the details, but I suspect you should work with the integral of φ over Xm-Xn for m>n. This -> 0 as m and n become infinite. Since |f| ≤ φ, the integral of f over the same domain will also -> 0 under the same condition.

3. Oct 17, 2014

### DavideGenoa

Thank you so much!!!