Littlewood’s three principles (2)

by tzy9393

The second principle asserts that every function is nearly continuous. Now we investigate the meaning of “nearly” in terms of mathematics.

Theorem 2 (Lusin) Suppose f is measurable and finite valued on E with E of finite measure. Then for every ${\epsilon>0}$ there exists a closed set ${F_{\epsilon}}$ with

$\displaystyle F_{\epsilon}\subseteq E\;\mathrm{and}\;m(E-F_{\epsilon})\leq\epsilon$

and such that ${f|_{F_{\epsilon}}}$ is continuous.

Proof: Since by assumption, ${f}$ is measurable and finite valued on E, we can find a sequence of step functions, ${f_{n}}$ such that ${f_{n}\rightarrow f}$ almost everywhere. Note that each ${f_{n}}$ is a step function which means it is continuous almost everywhere as it contains only at most countably many discontinuities which have Lebesgue measure 0. So without loss of generality, say for each ${n}$ we can find sets ${E_{n}}$ with ${m(E_{n})\leq\frac{1}{2^{n}}}$ so that ${f_{n}}$ is continuous outside ${E_{n}}$ . Now to proceed, we need something extra which is obtained from the third littlewood’s principle, i.e., Egrovff’s theorem. It enables us to find a set ${A_{\frac{\epsilon}{3}}}$ with ${m(E-A_{\frac{\epsilon}{3}})\leq\frac{\epsilon}{3}}$ such that ${f_{n}\rightarrow f}$ uniformly on ${A_{\frac{\epsilon}{3}}}$ . Then for the integer N such that ${\sum_{n\geq N}\frac{1}{2^{n}}\leq\frac{\epsilon}{3}}$ , we consider

$\displaystyle F':=A_{\frac{\epsilon}{3}}-\bigcup_{n\geq N} E_{n}$

For every ${n\geq N}$ , the step function ${f_{n}}$ is continuous on ${F'}$ ; therefore, the uniform limit ${f}$ is also continous on ${F'}$ . To complete the proof, notice that ${F'}$ is also measurable so we can approximate from below by a closed set ${F_{\epsilon}}$ such that ${m(F'-F_{\epsilon})\leq\frac{\epsilon}{3}}$ which gives us the desired ${m(E-F_{\epsilon})\leq\epsilon}$ .