Does $f(x)=\boldsymbol{1}_{(-9,3]}-2\cdot\boldsymbol{1}_{[11,\infty)}$ belong to $\mathcal{L}^{\infty}(\mu)$ with $\mu$ being the Lebesgue measure? Consider the function $f:\mathbb{R}\rightarrow\mathbb{R}$,
$$\tag{1}
f(x)=\boldsymbol{1}_{(-9,3]}-2\cdot\boldsymbol{1}_{[11,\infty)}
$$
I want to know if the function belongs to the function space $\mathcal{L}^{\infty}(\mu)$ with $\mu$ being the Lebesgue measure. My lecture book defines this function space as
$$\tag{2}
\mathcal{L}^{\infty}(\mu):=\{u: X \rightarrow \mathbb{R}: u \in \mathcal{M}(\mathscr{A}), \exists c>0, \mu\{|u| \geqslant c\}=0\}
$$
In an answer below, it was stated that $\{|u| \geqslant c\}=\{x\in\mathbb{R}:u(x)\ge c\}$. If we choose $c>2$, then $\{|u| \geqslant c\}=\emptyset$ and $\mu\{|u| \geqslant c\}=0$, which then implies that $f$ belongs to $\mathcal{L}^{\infty}(\mu)$. However, my lecture notes define the function space $\mathcal{L}^{1}(\mu)$ as
$$\tag{3}
\mathcal{L}^{1}(\mu):=\left\{f \in \mathcal{M} \mid \text { both } \int_{X} f^{+} d \mu <+\infty \text { and } \int_{X} f^{-} d \mu <+\infty\right\}
$$
where we have written the function as $f=f^+-f^-$. Consider then the Dirac delta measure concentrated at 12. We still have $\delta_{12}\{|f|\ge c\}=0$, so from definition $(2)$ the function should belong $\mathcal{L}^{\infty}(\delta_{12})$. But integrating $\boldsymbol{1}_{(-9,3]}$ and $2\cdot\boldsymbol{1}_{[11,\infty)}$ with respect to $\delta_{12}$ gives $0$ and $2$, so according to the definition in eq. $(3)$, the function also belongs to $\mathcal{L}^{1}(\delta_{12})$?
 A: You should interpret in words that $L^\infty$ is the set of all functions whose absolute value is bounded for all $x$ except maybe for a set of zero measure. For example we would call the extended real valued function $f(x)\begin{cases} 1\,,x\in\Bbb{R\setminus N}\\ \infty\,,x\in\Bbb{N}\end{cases}$ as a $L^\infty$ function. You can construct many such examples where f is unbounded as a whole but is bounded but for a set of measure zero.
$\{|u|\geq c\}=\{x\in\Bbb{R}:u(x)\geq c\} $.
So does there exist any such $x$ such that $|u(x)|\geq 4$ ?
If no then what can you say about the measure of the set  $\{|u|\geq 4\}$ ?
Now can you see why it belongs to $L^{\infty}(\mu)$?
Now can you conclude about finite linear combination of indicator functions? .
Dirac Measure is a finite measure. For any finite measure $\mu$ you have $L^{\infty}(\mu)\subset L^{1}(\mu)$ .
This is simple enough to prove .
as $f\in L^{\infty}(\mu)$ there exist $0<M<\infty$ such that $\mu(\{|f|\geq M\})=0$. Call the set $\{x\in X :|f(x)|\geq M\}=A$
Then You have $$\int_{X}|f|\,d\mu = \int_{A}|f|\,d\mu + \int_{X\setminus A} |f|\,d\mu= \int_{X\setminus A} |f|\,d\mu\leq \int_{X\setminus A} M\,d\mu\leq M\cdot \mu(X) <\infty$$.
Note that $\int_{A}|f|\,d\mu=0$ as $\mu(A)=0$ and this is a convention that is followed. Integral over sets of $0$ measure is taken to be $0$.
Also You are defining $L^{1}(\mu)$ as the set of functions for which integrals of $f^+$ and $f^-$ exist and are finite. Well that is good enough but as I said , a function is integrable (i.e . integrals of $f^+$ and $f^-$ exists and do so finitely if and only if the integral of $|f|=f^{+}+f^{-}$ exists and is finite . That is a function is integrable in the Lebesgue sense if and only if it is absolutely integrable($|f|$ is integrable). This is not true for improper Riemann integrals and indeed you have $\frac{\sin(x)}{x}\mathbf{1}_{(0,\infty)}$ is not integrable in the Lebesgue sense whereas we all know that this is the Dirichlet Integral and it's value is $\frac{\pi}{2}$.
Note 2 The result $L^{\infty}\subset L^{1}$ is false for non-finite measure spaces . For example in the case of the real line with Lebesgue measure , the function $1$ , i.e $\mathbf{1}_{\Bbb{R}}$ is $L^{\infty}(\lambda)$ as you can clearly see but it is not $L^{1}(\lambda)$.
Speaking more generally $L^{q}(\mu)\subset L^{p}(\mu)$ if $1\leq p\leq q \leq\infty$ if the measure is finite . A proof of this can be found in any good measure theory book. See Royden page 142 for instance.
And you can find easy examples of $L^{1}$ functions which are not $L^{2}$ . For example in $(0,1)$ you have $\frac{1}{\sqrt{x}}$ is $L^{1}$ but not $L^{2}$.
I suggest you study $L^{p}$ spaces more and try to come up with your own examples and counter examples to make your concept strong.
But these notions need not hold for infinite measures. There are $L^{2}$ functions which are not $L^{1}$, for example $\frac{1}{x}$ in $[1,\infty)$
with Lebesgue measure is $L^{2}$ but not $L^{1}$.
