The Unapologetic Mathematician

Mathematics for the interested outsider

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)<c if and only if f_n(x)<c for some n. That is, we can write

\displaystyle\{x\in X\vert g(x)<c\}=\bigcup\limits_{n=1}^\infty\{x\in X\vert f_n(x)<c\}

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


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  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 […]

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  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

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