Exercise 1.3.25 (Littlewood-like principles)

Fix an arbitrary $𝜖>0$.

(i)

By Theorem 1.3.20 (Approximation of absolutely integrable functions) we can find a compactly supported function $g$ with $\parallel f-g{\text{∥}}_{L^1}\le 𝜖$. Since $g$ is compactly supported, we can find an $R>0$ large enough so that $g(x)=0$ whenever $x\in {ℝ}^d∕B(0,R)$. With this we obtain $$𝜖\ge {\int \; <!--\; nolimits\; -->}_{{ℝ}^d}|f-g|={\int \; <!--\; nolimits\; -->}_{B(0,R)}|f-g|+{\int \; <!--\; nolimits\; -->}_{{ℝ}^d∕B(0,R)}|f-g|={\int \; <!--\; nolimits\; -->}_{B(0,R)}|f-g|+{\int \; <!--\; nolimits\; -->}_{{ℝ}^d∕B(0,R)}\left|f\right|\ge {\int \; <!--\; nolimits\; -->}_{{ℝ}^d∕B(0,R)}\left|f\right|.$$

(ii)

By Exercise 1.3.23 we can find a Lebesgue measurable set $A\subseteq {ℝ}^d$ such that $m(A)\le 𝜖∕2$ and $f$ is continuous on ${ℝ}^d∕E$. We can enlarge $A$ to an open set $U$ by expending another $𝜖∕2$, i.e., we can find an open set $U\subseteq {ℝ}^d$ such that $m(U)\le m(A)+𝜖∕2\le 𝜖$ and $f$ is continuous on ${ℝ}^d∕U$.

Now pick an arbitrary bounded set $B(0,R)$. Notice that the set $B(0,R)∕U=B(0,R)\cap U^c$ is both closed and bounded. By Heine-Borel Theorem (cf. Analyis II, Theorem 1.5.7) our set $B(0,R)∕U$ is compact. Thus, our continuous function $f$ must become uniformly continuous when restricted to $B(0,R)∕U$ (cf. Analysis II, Theorem 2.3.5). As such, it must be bounded on this set since uniformly continuous functions are bounded.

(for the last assertion: if we argue by contradiction, then we can construct a sequence ${(x_n)}_{n\in \mathbb{N}}$ for which $f_n(x_n)>n$ for $n\in \mathbb{N}$, i.e., $(f_n{(x_n)}_{n\in \mathbb{N}}\to \infty$. But ${(x_n)}_{n\in \mathbb{N}}$ has a convergent subsequence, and we have ${(x_{n_i})}_{i\in \mathbb{N}}\to x_0$ yet $f(x_{n_i})>n_i$ which contradicts to $d(x_n,x_m)\le \delta ⟹d(f(x_n),f(x_m))\le 𝜖$).