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.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s