Littlewood’s first principle says that each measurable set E of finite measure is nearly a union of finitely many open intervals. However, if the G for which the symmetric difference with E is less than epsilon, and if we add the condition that E must be a subset of G Littlewood's principle fails when E is an unbounded set.
I find these counterexamples to be very instructive for exploring the boundaries these principles define. Would anyone be able to point me in the direction of similar counterexamples for Littlewood's second and third principles (Luzin's and Egorov's theorems respectively)?
$\endgroup$1 Answer
$\begingroup$The statement of Egorov's theorem is as follows:
Let $(X, \Sigma, \mu)$ be a measure space. Let $D \in \Sigma$ be such that $\mu (D) < \infty$. Let $(f_n)_{n \in \mathbb{N}}, f_n: D \to \mathbb{R}$ be a sequence of $\Sigma$-measurable functions. Suppose that $f_n$ converges a.e. to $f$, for some $\Sigma$-measurable function $f: D \to \mathbb{R}$. Then $f_n$ converges uniformly a.e. to $f$.
Here there are two important hypothesis without which the theorem will not hold.
$\mu(D) < \infty$. Counterexample: $f_n = \chi_{[n,n+1]}$ defined on the entire $\mathbb{R}$
The function $f$ must be finite, at least a.e. Counterexample: $f_n = n$ and $f = \infty$.
And the statement of Lusin's is
Let $f: \mathbb{R} \rightarrow \mathbb{R}$. Then TFAE
(a) $f$ is measurable.
(b)For every $\epsilon > 0$, there is $E \subset \mathbb{R}$ such that $\mu(E) < \epsilon$ and $f_{\vert \mathbb{R}\setminus E}$ is continuous.
The worst case example for this is $f = \chi_\mathbb{Q}$ which is no where continuous, where as $f_{\vert \mathbb{Q}^c}$ is continuous.
Hope this helps
$\endgroup$ 1
