The space $L^\infty$ consists of functions which are essentially bounded. Generally speaking, extended valued functions may take on the values of $\pm \infty $, in which case if we look at the sup norm of such functions, we may have $\sup_{x \in{X}}|f(x)|=\infty$. If we want to examine the bounds of a function almost everywhere, then we define the $\textit{essential sup}$ of $f$: $$\text{esssup}(f) = \inf\{M>0 : |f(x)| \leq M \quad a.e.\} = ||f||_\infty $$ This is a generalization of the usual sup norm and is the same as the sup norm for bounded functions such as $C[a,b]$. Some examples:
1. $f(x) = +\infty \chi_\mathbb{Q}$. Although $f$ takes on the value of $+\infty$ infinitely many times, it does so on a set of measure 0. Thus $||f||_\infty = 0$
2. $f(x) = -\log(x)\chi_{[0,1]}$. Then, $\int_0^1 \log^p(x) = \Gamma(p+1)$ and so $f\in{L^p[0,1]}$ for any finite $p$. However, $\forall M>0$, there exists an interval, namely $[0,e^{-M})$ which has positive measure, $e^{-M}$, where $f>M$. Thus, $f \not \in{L^\infty[0,1]}$
In general, we can make an inclusion relation between $L^p(E)$ and $L^q(E)$ only when $\mu(E)<\infty$.
$\textit{Lemma}: ||\cdot||_\infty$ is a norm.
(i) $f=0$ a.e. $\Longleftrightarrow$ $|f| \leq 0$ a.e. $\Longleftrightarrow$ $||f||_\infty = 0$.
(ii) Let $\alpha \in{\mathbb{R}}$ (or $\alpha\in{F}$...whatever underlying field of scalars our functions map to). Then, $||\alpha f||_\infty = \inf\{M \geq 0: |\alpha f(x)| \leq M \text{ a.e. x} \} = \inf\{M \geq 0: |\alpha| \cdot | f(x)| \leq M \text{ a.e. x} \} = |\alpha | \cdot \inf\{M \geq 0: | f(x)| \leq M \text{ a.e. x} \} = |\alpha| \cdot ||f||_\infty$
(iii) $||f+g||_\infty = \inf\{M \geq 0 : |f(x)+g(x)| \leq M \text{ a.e. x} \} \leq \inf\{M_1 \geq 0 : |f(x)| \leq M_1 \text{ a.e. x} \} + \inf\{M_2 \geq 0 : |g(x)| \leq M_2 \text{ a.e. x} \} = ||f||_\infty + ||g||_\infty $
We now wish to show that $L^\infty(E \subseteq X,\mu) = \{f: E\in{\mathcal{M}} \longrightarrow [-\infty,\infty]: \quad ||f||_\infty < \infty \}$ together with the norm $|| \cdot ||_\infty$ is complete: If $||f_n-f_m||_\infty \longrightarrow 0$ as $n,m \longrightarrow \infty$ then $||f_n-f||_\infty \longrightarrow 0$ where $f\in{L^\infty}$
$\textit{proof}$ :
Let $\{f_n\}$ be a Cauchy sequence with respect to $||\cdot||_\infty$. Define $$A_n = \{x \in{E} : |f_n(x)| > ||f_n||_\infty \} $$ $$B_{mn} = \{x\in{E}: |f_n(x)-f_m(x)| > ||f_n-f_m||_\infty \}$$ By definition, $\mu(A_n) = \mu(B_{mn}) = 0$, so $\mu(A_n \cup B_{mn}) \leq \mu(A_n) + \mu(B_{mn}) = 0+0 = 0$.
For any fixed $ x \in{(A_n \cup B_{mn})^c}$, $\{|f_n(x)|\}_{n=1}^\infty $ and $\{|f_n(x)-f_m(x)|\}_{n,m=1}^\infty$ are bounded sequences of non-negative real numbers. $\{|f_n(x)-f_m(x)|\}_{n,m=1}^\infty$ is a Cauchy sequence of real numbers and since $(\mathbb{R},| \cdot|)$ is complete, this sequence is convergent. We can see this as follows: On $(A_n \cup B_{mn})^c$, the $L^\infty$ norm is the same as the sup norm $$|f_n(x) - f_m(x)| \leq ||f_n - f_m||_\infty = \sup_{x\in{(A_n\cup B{mn})^c}}|f_n(x)-f_m(x)| \longrightarrow 0 \text{ as } n,m \longrightarrow \infty$$ Convergence in the sup norm is equivalent to uniform convergence. Letting $m \longrightarrow \infty$, $$|f_n(x)-f(x)| \leq \sup_{x\in{(A_n\cup B{mn})^c}}|f_n(x)-f(x)| \longrightarrow 0 \text{ as } n \longrightarrow \infty$$ So, $f_n(x) \longrightarrow f(x)$ uniformly $\forall x\in{(A_n \cup B_{mn})^c}$ thus forcing $f$ to be bounded a.e.
Case 2: $p=1$
Consider the following examples:
1. $f_n = \frac{1}{n}$. Then $f_n \longrightarrow 0$ (in fact, uniformly) on all of $\mathbb{R}$, yet $\int_\mathbb{R} f_n = \infty$ $\forall n \in{\mathbb{N}}$ while $\int_\mathbb{R}0 = 0$. This shows us that pointwise (or even uniform) convergence does not imply convergence in the $L^1$ norm.
2. Let $f_1 = \chi_{[0,1]}$, $f_2 = \chi_{[0,\frac{1}{2}]}$, $f_3 = \chi_{[\frac{1}{2},1]}$, $f_4 = \chi_{[0,\frac{1}{3}]}$, $f_5 = \chi_{[\frac{1}{3},\frac{2}{3}]}$, $f_6 = \chi_{[\frac{2}{3},1]}$, $f_7 = \chi_{[0,\frac{1}{4}]}$, ...
So, $||f_n||_1 \longrightarrow 0$ yet $\{f_n\}$ does not have a pointwise limit since the $f_n$ continue to "sweep" back from 0 to 1. Indeed, $\forall x\in{[0,1]}$, $\exists N_x \in{\mathbb{N}}$ s.t. $f_N(x) = 1$.
$\{f_n\}$ still manages to be Cauchy in $L^1$ since, for $n,m \in{\mathbb{N}}$, $||f_n-f_m||_1 \leq \frac{1}{n} + \frac{1}{m} \longrightarrow 0$ as $n,m \longrightarrow \infty$
Despite this, we can still manage to find a subsequence $\{f_{n_k} \}$ of $\{f_n\}$ that converges pointwise a.e. to $f=0$. For instance, the subsequence $f_1, f_2,f_4,f_7,...$ is s.t. $f_{n_k}= \chi_{[0,\frac{1}{k}]} \longrightarrow 0$ a.e. For an even faster converging subsequence, choose $f_2,f_7,...$. In this case, $||f_{n_{k+1}}-f_{n_k}||_1 = \int |\chi_{[0,\frac{1}{2^k}]}-\chi_{[0,\frac{1}{2^{k-1}}]}| = \int \chi_{[0,\frac{1}{2^k}]} = \frac{1}{2^k}$. These types of subsequences which converge pointwise as well as in the $L^1$ norm are called $\textit{rapidly Cauchy}$ subsequences.
$\textit{proof}$
Let $\{f_n\}_{n=1}^\infty$ be Cauchy in $L^1$. We wish to show that $||f_n-f||_1 \longrightarrow 0$ and that $f\in{L^1}$. Choose a subsequence $\{f_{n_k}\}$ that is "rapidly Cauchy" i.e. $$||f_{n_{k+1}} - f_{n_k}||_1 \leq \frac{1}{2^k} \quad \forall k\in{\mathbb{N}}$$ Such a subsequence exists from assumption that $\{f_n\}$ is Cauchy. By induction, for any $k \in{\mathbb{N}}$, we may choose an $n_k$ (now depends on $k$) sufficiently large s.t. the above inequality holds. Since $f_{n_k} $ are integrable, they are measurable. If we can show that $|f_{n_k}-f|$ is dominated by an $L^1$ function and $f_{n_k}-f \longrightarrow 0$ a.e. then the dominated convergence theorem will imply $||f_{n_k}-f||_1 \longrightarrow 0$. Since $\{f_n\}$ is assumed to be Cauchy, $||f_{n_k}-f_n|| \longrightarrow 0$. An "$\frac{\epsilon}{2}$" argument and the triangle inequality of the $L^1$ norm give $$||f_n-f||_1 = ||f_n-f_{n_k}+f_{n_k}-f||_1 \leq ||f_n-f_{n_k}||_1 + ||f_{n_k}-f||_1 \overset{n\rightarrow \infty}{\longrightarrow} 0$$ The key property of the extracted subsequence $\{f_{n_k}\}$ suggests that we should define $f$ in terms of a telescoping series. Let $$f(x) = f_{n_1}(x) + \sum_{k=1}^\infty(f_{n_{k+1}}(x)-f_{n_k}(x))$$ By the generalized triangle inequality and the continuity of the absolute value function, $$|f| = \left|f_{n_1} + \sum_{k=1}^\infty(f_{n_{k+1}}-f_{n_k})\right| \leq |f_{n_1}| + \left|\lim_{N \rightarrow \infty } \sum_{k=1}^N(f_{n_{k+1}}-f_{n_k})\right| = |f_{n_1}| + \lim_{N \rightarrow \infty }\left| \sum_{k=1}^N(f_{n_{k+1}}-f_{n_k})\right|$$ $$ \leq |f_{n_1}| + \lim_{N \rightarrow \infty } \sum_{k=1}^N\left|f_{n_{k+1}}-f_{n_k}\right| = |f_{n_1}| + \sum_{k=1}^\infty|f_{n_{k+1}}-f_{n_k}|=g$$ The above sequence of partial sums is non-negative and increasing, so apriori, the limit may be $\infty$. By the monotinicity and linearity of the Lebesgue integral, $$||f||_1 = \int|f| \leq \int|f_{n_1}| + \int \sum_{k=1}^\infty |f_{n_{k+1}} - f_{n_k}| = \int |f_{n_1}| + \sum_{k=1}^\infty \int |f_{n_{k+1}} - f_{n_k}|$$ where the justification for the interchange of countable sum and integral comes from noting that $0 \leq \sum_{k=1}^N |f_{n_{k+1}} - f_{n_k}| \nearrow \sum_{k=1}^\infty |f_{n_{k+1}} - f_{n_k}|$ and applying the monotone convergence theorem. $$=||f_{n_1}||_1 + \sum_{k=1}^\infty ||f_{n_{k+1}}-f_{n_k}||_1 \leq ||f_{n_1}||_1 + \sum_{k=1}^\infty \frac{1}{2^k} = ||f_{n_1}||_1 + 1 < \infty$$ So, $|f| \leq g\in{L^1} $ hence $ f\in{L^1}$. In particular, the series defining $f$ converges a.e. and since the partial sums are exactly $f_{n_k}$: $$f(x) = f_{n_1}(x) + \lim_{N \rightarrow \infty} \sum_{k=1}^N (f_{n_{k+1}}(x)-f_{n_k}(x)) = \lim_{N \rightarrow \infty}f_{n_N}(x) \quad \text{a.e. } x$$ So, $\{f_{n_k}-f\}$ is a sequence of $L^1$ (hence measurable) functions and $f_{n_k}-f \longrightarrow 0$ a.e. Also, $$|f_{n_N}| = |f_{n_1}| + \sum_{k=1}^N|f_{n_{k+1}}-f_{n_k}| \leq |f_{n_1}| + \sum_{k=1}^\infty|f_{n_{k+1}}-f_{n_k}|=g$$ By the triangle inequality, $|f_{n_k}-f| \leq |f|+|f_{n_k}| \leq g+g=2g \in{L^1}$. Applying the dominated convergence theorem gives $||f_{n_k}-f||_1 \longrightarrow 0$ The case for $1
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