# Measurability and Lebesgue integral of complex functions

1. Oct 18, 2014

### DavideGenoa

Hi, friends! I think that it is correct to say that, given a measure space $X$:
• if $f:X\to\mathbb{R}$, $\tilde{f}:X\to\mathbb{C}$ and $\forall x\in X\quad f(x)=\tilde{f}(x)$, then $f$ is measurable if and only if $\tilde{f}$ is;
• $f:X\to\mathbb{C}$ is Lebesgue integrable if and only if both $\text{Re}f:X\to\mathbb{R}$ and $\text{Im}f:X\to\mathbb{R}$ are; in that case $\int_X f(x)d\mu=\int_X \text{Re}f(x)d\mu+i\int_X \text{Im}f(x)d\mu$.
Am I right? $\infty$ thanks!

2. Oct 18, 2014

### mathman

Second statement looks OK.

First statement is confusing - I can't understand the relationship and/or difference between these two functions.

3. Oct 18, 2014

### DavideGenoa

I mean: the Borel algebra of $\mathbb{R}$ and $\mathbb{C}$ are different, but I have convinced myself of that equivalence because I think that the Borel sets of $\mathbb{R}$ are Borelian as subsets of $\mathbb{C}$ and any Borel set belonging to $\mathbb{C}$ contained in $\mathbb{R}\subset\mathbb{C}$ is a Borel set considered as a subset of $\mathbb{R}$; therefore I think that our function, which can be considered as a complex or real function defined on the measure space $X$ with measure $\mu$, is measurable with respect to the Borel algebra of $\mathbb{R}$ (in the sense that the counterimage of any Borel set is $\mu$-measurable), i.e. as a real function, if and only if it is with respect to the Borel algebra of $\mathbb{C}$, i.e. as a complex function.
Thank you so much again!

4. Oct 19, 2014

### mathman

Since the domain (X) is the same real line in either case, the issue you are raising is not relevant. X is the measure space. The range of the function does not come into play.

5. Oct 20, 2014

### DavideGenoa

Thank you so much!