Pratt's Lemma

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Theorem

Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

Let $\left({g_n}\right)_{n \in \N}$, $\left({G_n}\right)_{n \in \N}$ and $\left({f_n}\right)_{n \in \N}$ be sequences of $\mu$-integrable functions.

Suppose that the pointwise limits $\displaystyle f := \lim_{n \to \infty} f_n$, $\displaystyle g := \lim_{n \to \infty} g_n$ and $\displaystyle G := \lim_{n \to \infty} G_n$ exist, and that $g$ and $G$ are $\mu$-integrable.

Suppose further that, for all $x \in X$ and $n \in \N$:

$g_n \left({x}\right) \le f_n \left({x}\right) \le G_n \left({x}\right)$

Finally, suppose the following hold:

$\displaystyle \lim_{n \to \infty} \int g_n \, \mathrm d \mu = \int g \, \mathrm d \mu$
$\displaystyle \lim_{n \to \infty} \int G_n \, \mathrm d \mu = \int G \, \mathrm d \mu$


Then:

$\displaystyle \lim_{n \to \infty} \int f_n \, \mathrm d \mu = \int f \, \mathrm d \mu$

and the latter is finite.


Proof


Source of Name

This entry was named for John Winsor Pratt.


Sources