Given a measurable space $X$ equipped with a measure $\mu$, a function $f : X \to \mathbb{C}$ which is defined almost everywhere (that is, up to a set of measure $0$) is said to be an element of $L^1$ if
$$\int_X |f| d\mu < \infty$$
More properly, we have to identify functions which are equal almost everywhere, so the elements of the Lebesgue space $L^1$ are really equivalence classes of functions under the relation of being almost everywhere equal - but this is a technical note.
Practically speaking, a real or complex valued measurable function on the real line with respect to Lebesgue measure is an element of $L^1$ if $$\int_{-\infty}^{\infty} |f(x)| dx < \infty$$
So a function like $\chi_{[0,1]}$ which is $1$ on $[0,1]$ and $0$ outside is in $L^1$, as is $e^{-x^2}$.
If $f$ is continuous enough, this coincides with the usual Riemann integral. Now it's fairly easy to prove that $$\int_{\mathbb{R}} |\sin x| dx$$ can't be finite, so $\sin x \notin L^1(\mathbb{R})$. In some sense, $L^1$ functions have to decay to $0$ at $\pm \infty$: In fact, one way to think of $L^1$ is that it's the completion of
$$C_C = \{\text{continuous functions supported on a compact set}\}$$
under the metric induced by integration (again, with slight technical caveats).
So in short, ignoring the technical definitions, $L^1$ functions are exactly those functions which have small enough spikes and decay fast enough that $\int |f| < \infty$.
Best Answer
For $f :\Omega \to \mathbb R $ measurable, you can define its essential supremum by $$ E(f) := \inf_{\substack{U: \mu(U) = 0}} \sup_{x\in \Omega \setminus U} |f(x)|$$
Suppose that $f=0$ a.e. Let the null set where $f\neq 0$ be $U_0$. Then $$ 0 \le E(f) \le \sup_{x\in \Omega\setminus U_0} |f(x)| = 0$$
so $E(f) = 0$.
Conversely, if $E(f)=0$, Then for any $\epsilon>0$ there is a null set $U$ such that $$\sup_{x\in\Omega\setminus U}|f(x)| < \epsilon $$ Thus, $|f|<\epsilon$ a.e. for every $\epsilon>0$, and $$\mu(|f|>0) = \mu \left(\bigcup_{n=1}^\infty \{ |f|>1/n \}\right) \le \sum_{n=1}^\infty \mu(|f|>1/n) = \sum_{n=1}^\infty 0 = 0$$ so $f=0$ a.e. This proves that $E(f)=0 \iff f=0$ a.e.
Now, if $f=g$ a.e. then $E(f-g)=0$, and so taking sups and then infs of $|f(x) \le |g(x)|+|f(x) -g(x)|$ gives $$ E(f) \le E(g) + E(f-g) = E(g)$$ and similarly $E(g)\le E(f)$, so $E(f) = E(g)$.
Hence, if you take the set $$ \mathcal L^\infty(\Omega):= \{ f:\Omega \to \mathbb R \text{ measurable} : E(f) <\infty \}$$ and quotient out by $f\sim g \iff f=g$ a.e., $E([f]):=E(f)$ is well defined on the equivalence classes $[f]\in L^\infty(\Omega):=\mathcal L^\infty(\Omega)/\sim $. One can then go through the standard arguments to check that $E$ is a norm. The above takes care of the hardest step, namely $E([f])=0$ iff $E(f) = 0$, iff $f=0$ a.e., iff $[f]=[0]$.
To match with the definition in Wikipedia, note that for each $\epsilon$, there is a null set $U$ such that $|f|<E(f)+\epsilon$ outside $U$, so that $\|f\|_\infty \le E(f)$ (with $\|f\|_{\infty}$ as defined in Wikipedia). Conversely, if $|f(x)|<C$ a.e., Then $E(f) \le C$. Taking infimums over allowable $C$s gives $E(f) \le \|f\|_{\infty}$, so $E(f) = \|f\|_{\infty}$.