The Unapologetic Mathematician

Mathematics for the interested outsider

Integrable Functions

Now we can define what it means for a general real-valued function (not just a simple function) to be integrable: a function f is integrable if there is a mean Cauchy sequence of integrable simple functions \{f_n\} which converges in measure to f. We then define the integral of f to be the limit

\displaystyle\int f(x)\,d\mu(x)=\int f\,d\mu=\lim_{n\to\infty}\int f_n\,d\mu

But how do we know that this doesn’t depend on the sequence \{f_n\}?

We recall that we defined

\displaystyle N(f)=\{x\in X\vert f(x)\neq0\}

which must be measurable for any measurable function f. This is the only part of the space that matters when it comes to integrating f; clearly we can see that

\displaystyle\int f\,d\mu=\int\limits_{N(f)}f\,d\mu

since f is zero everywhere outside N(f).

Now, if both \{f_n\} and \{g_n\} converge in measure to f, then we can define E to be the (countable) union of all the N(f_n) and N(g_n). Just as clearly, we can see that

\displaystyle\begin{aligned}\int f_n\,d\mu&=\int\limits_Ef_n\,d\mu=\nu_n(E)\\\int g_n\,d\mu&=\int\limits_Eg_n\,d\mu=\lambda_n(E)\end{aligned}

where \nu_n is the indefinite integral of f_n, and \lambda_n is the indefinite integral of g_n. Then if we use \{f_n\} to define the integral of f we get

\displaystyle\lim\limits_{n\to\infty}\int f_n\,d\mu=\lim\limits_{n\to\infty}\nu_n(E)=\nu(E)

while if we use \{g_n\} we get

\displaystyle\lim\limits_{n\to\infty}\int g_n\,d\mu=\lim\limits_{n\to\infty}\lambda_n(E)=\lambda(E)

But we know that since \{f_n\} and \{g_n\} both converge in measure to the same function, the limiting set functions \nu and \lambda coincide, and thus \nu(E)=\lambda(E). The value of the integral, then, doesn’t depend on the sequence of integrable simple functions!

June 2, 2010 - Posted by | Analysis, Measure Theory


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