# Lebesgue Integration Part 4- Complex Functions

If we have a complex measurable $f = u+iv$, where $u,v$ are real and measurable, then we define the Lebesgue integral of $f$ over measurable set $E\subset X$ as follows:

$\displaystyle\int_E f \ d\mu = \displaystyle\int_E u^+ \ d\mu - \displaystyle\int_E u^- \ d\mu+i\displaystyle\int_E v^+ \ d\mu - i\displaystyle\int_E v^- \ d\mu$,

where for a real function $g$, $g^+$ and $g^-$ denote $\max(g,0)$ and $-\min(-g,0)$ respectively so that $g = g^+ - g^-$. We say that $f\in L^1(\mu)$, that is, $f$ is Lebesgue-integrable, if $\int_X |f| \ d\mu<\infty$. It is easy to verify the usual additive and scalar multiplicative properties of the complex Lebesgue integral.

Now that we can work with arbitrary complex measurable functions, we get the following inequality:

Result 12: $|\int_X f \ d\mu|\le \int_X|f| \ d\mu$.

Proof: This makes intuitive sense as it is essentially a triangle inequality. Think of $\int_X f \ d\mu$ as a complex number $z$, and let $\alpha$ be a “straightening rotation factor” so that $\alpha z = |z|$ and $|\alpha| =1$; in other words, $\alpha z$ is simply $z$ rotated onto the real line. Then

$\left|\displaystyle\int_X f \ d\mu\right| = \alpha\displaystyle\int_X f \ d\mu = \displaystyle\int_X \alpha f \ d\mu$.

Note, however, that these values are all real, so $\int_X \alpha f \ d\mu = \int_X \Re(\alpha f) \ d\mu \le \int_X |f| \ d\mu$, and we are done. We will now use this triangle inequality to prove a result that gives sufficient conditions for the Lebesgue-integrability of the limit of a sequence of complex, measurable functions and further states that the integrals of those functions approach the integral of their limit.

Result 13: Given complex measurable functions $\{f_n\}$ which converge to some $f$. Say they are “dominated” by some Lebesgue-integrable $g$ so that $|f_n|\le g$ over the entire space $X$ for all $n$. Then $f$ is Lebesgue-integrable and $\int_X f \ d\mu = \lim_{n\to\infty}\int_X f_n \ d\mu$.

Proof: We have essentially already proven that $f$ is measurable, and because $|f|$ is bounded above by $g$, $f$ is certainly Lebesgue-integrable. To prove the second result, it suffices by Result 12 to prove that $\lim_{n\to\infty} \int_X |f_n-f| \ d\mu =0$. Consider the distance between $f_n$ and $f$. This is bounded above by $2g$, so consider $2g-|f_n-f|$. Fatou’s Lemma tells us that

$\displaystyle\int_X \liminf_{n\to\infty} 2g-|f_n-f| \ d\mu \le \liminf_{n\to\infty}\displaystyle\int_X 2g-|f_n-f| \ d\mu$.

Because $\{f_n\}$ approaches $2g$, the left side is $\int_X 2g \ d\mu$. We can split up the right side into $\int_X 2g \ d\mu-\limsup_{n\to\infty}\int_X |f_n-f| \ d\mu$, so $\limsup_{n\to\infty}\int_X |f_n-f| \ d\mu \le 0$. But the limit is at most equal to the upper limit, and certainly $\lim_{n\to\infty}\int_X |f_n-f| \ d\mu\ge$, so we are done.