Problem 3

Exercise 1.3 (Exact Egorov's Theorem)   Suppose $f_n\to f$ almost everywhere. Find $E^k$ such $f_n\to f$ uniformly on $E^k$ and

$$\mu\left(\mathbb{R}\setminus\bigcup_{k=1}^\infty E^k\right) = 0.$$ (1.8)

Proof. For each positive integer $k$ and for each integer $m$, select

$$E_m^k \subseteq [m,m+1]$$ (1.9)

such that $f_n\to f$ uniformly on $E_m^k$ and $\mu([m,m+1]\setminus E_m^k)<1/k2^{\vert m\vert}$. Then define

$$E^k = \bigcup_{m=-k}^k E_m^k$$ (1.10)

To see $f_n\to f$ uniformly on $E^k$, let $m$ be fixed and $\epsilon>0$. By the uniform convergence of $f_n\to f$ on $E_m^k$, select $N_m$ such that $n\geq N_m$ implies $\Vert f_n-f\Vert _{E_m^k}<\epsilon$. Let $N=\min\{-N_k,\dots,N_k\}$ to see that the convergence is uniform on the union $E^k$ also.

Break apart the complement

$$\mathbb{R}\setminus E^k = \left(\bigcup_{j=-\infty}^\infty [j,j+1]\right)\setminus \bigcup_{m=-k}^k E_m^k$$ (1.11)

$$= \bigcup_{j=-\infty}^\infty \left([j,j+1]\setminus \bigcup_{m=-k}^k E_m^k\right)$$ (1.12)

$$= \left(\bigcup_{j=-k}^k [j,j+1]\setminus E_j^k\right)\cup\bigcup_{\vert j\vert>k} [j,j+1]$$ (1.13)

The finite union can be measured by hand:

$$\mu\left(\bigcup_{j=-k}^k [j,j+1]\setminus E_j^k\right) = \sum_{j... ...mu([j,j+1]\setminus E_j^k) \leq \sum_{j=-\infty}^\infty 1/k2^{\vert j\vert}=3/k$$ (1.14)

To study the infinite union, label

$$U_k = \bigcup_{\vert j\vert>k} [j,j+1]$$ (1.15)

From the definition, we know that $U_1\supseteq U_2\supseteq\cdots$ and since no element in $\mathbb{R}$ is infinite, the intersection $\bigcap U_k$ is empty. Thence,

$$\mathbb{R}\setminus\bigcup_{k=1}^\infty E^k = \bigcap_{k=1}^\infty \mathbb{R}\setminus E^k$$ (1.16)

$$= \bigcap_{k=1}^\infty\left[ \left(\bigcup_{j=-k}^k [j,j+1]\setminus E_j^k\right)\cup\bigcup_{\vert j\vert>k} [j,j+1]\right]$$ (1.17)

$$= \left[\bigcap_{k=1}^\infty \left(\bigcup_{j=-k}^k [j,j+1]\setminus E_j^k\right)\right]\cup\left[ \bigcap_{k=1}^\infty U_k \right]$$ (1.18)

Therefore,

$$\mu\left(\mathbb{R}\setminus\bigcup_{k=1}^\infty E^k\right)<3/k \quad \forall k$$ (1.19)

which implies the measure of this set equals zero. $\qedsymbol$