# The Unapologetic Mathematician

## 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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