# The Unapologetic Mathematician

## Sequences of Measurable Functions

We let $\{f_n\}_{n=1}^\infty$ be a sequence of extended real-valued measurable functions on a measurable space $X$, and ask what we can say about limits of this sequence.

First of all, the function $g(x)=\inf\limits_{n\geq1}\{f_n(x)\}$ is measurable. The preimage $g^{-1}(\{-\infty\})$ is the union of the countable collection $\left\{f_n^{-1}(\{-\infty\})\right\}$, while the preimage $g^{-1}(\{\infty\})$ is the intersection of the countable collection $\left\{f_n^{-1}(\{\infty\})\right\}$. And so both of these sets are measurable, and we can restrict to the case of finite-valued functions.

So now let’s use our convenient condition. Given a real number $c$ we know that $g(x) if and only if $f_n(x) for some $n$. That is, we can write

$\displaystyle\{x\in X\vert g(x)

Each term on the right is measurable since each $f_n$ is a measurable function, and so the set on the left is measurable. Thus we conclude that $g$ is measurable as well.

Similarly, we find that the function $h(x)=\sup\limits_{n\geq1}\{f_n(x)\}=-\inf\limits_{n\geq1}\{-f_n(x)\}$ is measurable.

Now the functions

\displaystyle\begin{aligned}f^*(x)&=\limsup\limits_{n\to\infty}f_n(x)=\inf\limits_{n\geq1}\sup\limits_{m\geq n}f_m(x)\\f_*(x)&=\liminf\limits_{n\to\infty}f_n(x)=\sup\limits_{n\geq1}\inf\limits_{m\geq n}f_m(x)\end{aligned}

are also measurable. Indeed, in proving that $f^*$ is measurable we can use the exact same technique as above to prove that the inner supremum is measurable; it doesn’t really depend on the supremum starting at $1$ or higher. And then the outer infimum is exactly as before. Proving $f_*$ is measurable is similar.

Now we can talk about pointwise convergence of a sequence of measurable functions. That is, for a fixed point $x\in X$ we have the sequence $\{f_n(x)\}$ which has some limit superior $f^*(x)$ and some limit inferior $f_*(x)$. If these two coincide, then the sequence has a proper limit $\lim\limits_{n\to\infty}f_n(X)=f^*(x)=f_*(x)$. But one of our lemmas tells us that the set of points where any two measurable functions coincide has a nice property: $\{x\in X\vert f^*(x)=f_*(x)\}$ has a measurable intersection with every measurable set. And thus if we define the function $f(x)=\lim\limits_{n\to\infty}f_n(x)$ on this subspace of $X$ for which the limit exists, the resulting function is measurable.

May 10, 2010 - Posted by | Analysis, Measure Theory

1. […] functions! That is, given any measurable function we can find a sequence of simple functions converging pointwise to […]

Pingback by Simple and Elementary Functions « The Unapologetic Mathematician | May 11, 2010 | Reply

2. […] Convergence Almost Everywhere Okay, so let’s take our idea of almost everywhere and apply it to convergence of sequences of measurable functions. […]

Pingback by Convergence Almost Everywhere « The Unapologetic Mathematician | May 14, 2010 | Reply

3. […] sequence of measurable functions and concludes that is measurable (along with the inequality), but we already know that the limit inferior of a sequence of measurable functions is measurable, and so the integrable […]

Pingback by Fatou’s Lemma « The Unapologetic Mathematician | June 16, 2010 | Reply

4. I am reading simple measurable functions, but I couldn’t locate the convergence of simple measureable functions precisely , can anyone help me in that area?

Comment by manju | September 23, 2012 | Reply