Given an LTI system, the system is BIBO stable if and only if the impulse response is absolutely summable or absolutely integrable.

Let $LTI$ be the statement ``$H$ is an LTI system''. 

Let $BIBO$ be the statement ``$H$ is BIBO stable''.


Then the following statement is a tautology:
$$ LTI \wedge BIBO$$
$$ \iff $$
$$\lr{ \sum \limits_{n\in\Z}\abs{h(n)} \lt \infty}  \vee \lr{\int_{-\infty}^{\infty}\abs{h(t)}dt \lt \infty  }  $$

Here is the proof of the forward direction using summations:
\begin{proof}

First we are going to show $\sum \abs{h(n)} \lt \infty \Rightarrow BIBO$. Assume $\sum \limits_{n \in \Z} \abs{h(n)} \lt \infty$:
\begin{align*}
y(n) &= \sum \limits_{k\in\Z}h(k)x(n-k) \\
\abs{y(n)} &= \abs{\sum \limits_{k\in\Z}h(k)x(n-k)} \\
\abs{y(n)} &\leq \sum \limits_{k\in\Z}\abs{h(k)x(n-k)} \\
\abs{y(n)} &\leq \sum \limits_{k\in\Z}\abs{h(k)}\abs{x(n-k)} \\
\end{align*}

To make the input bounded, let $\abs{x(n)} \lt B_x$ $\forall n \in \Z$. Then

\begin{align*}
 \abs{y(n)} &\leq \sum \limits_{k\in\Z}\abs{h(k)}B_x \\
 \abs{y(n)} &\leq B_x\sum \limits_{k\in\Z}\abs{h(k)} \\
 \abs{y(n)} &\leq B_x\cdot k \quad \mbox{where } k = \sum \abs{h(n)} \lt \infty \\
 \abs{y(n)} &\leq B_y \\
\end{align*}

So given that an impulse response of a system is absolutely summable, when passed a bounded input signal, the system will always produce a bounded output signal. Therefore, any system whose impulse response is absolutely summable is BIBO stable.
\end{example}

\begin{example}
Given the system defined by $y(n) - \alpha y(n-1) = x(n)$, with zero initial conditions to make it an LTI system, for what values of $\alpha$ can this system be BIBO stable? \\

Solve this system in the forward direction $y(n) = \alpha y(n-1) + x(n)$:
\begin{align*}
y(n) &= x(n) + \alpha y(n-1) \\
y(n) &= x(n) + \alpha \lr{x(n-1) + \alpha y(n-2)} \\
y(n) &= x(n) + \alpha x(n-1) + \alpha^2 y(n-2) \\
y(n) &= x(n) + \alpha x(n-1) + \alpha^2\lr{x(n-2) + \alpha y(n-3)} \\
y(n) &= x(n) + \alpha x(n-1) + \alpha^2 x(n-2) + \alpha^3 y(n-3) \\
y(n) &= x(n) + \alpha x(n-1) + \alpha^2 x(n-2) + \alpha^3 x(n-3) + \cdots \\
y(n) &= \sum \limits_{k \in \Z_\oplus} \alpha^k x(n-k)
\end{align*}
This shows us that $h(n) = \begin{cases} 0 & n\lt0 \\ \alpha^n & n\geq 0 \end{cases} $, or $h(n) = \alpha^n u(n)$. Now we can determine which values for which the system is BIBO stable. We can look at the absolute sum of the impulse response $h(n)$:
$$ \sum \limits_{k \in \Z} \abs{h(n)} = \sum \limits_{k \in \Z_\oplus} \abs{\alpha^k} = \frac{1}{1-\abs{\alpha}} $$
But that last equality is true if and only if $\abs{\alpha} \lt 1$, thus our range of values for which the system is BIBO stable is $\alpha \in (-1,1)$.

\end{example}

It is important to note that you cannot assume that $H(\omega)$ exists to show that a system is BIBO stable, because that presumes BIBO stability.




So far, we showed that absolute summability or absolute integrability is a sufficient condition for BIBO stability. Lets show that it is also necessary. We want to show that BIBO stability implies absolute summability or absolute integrability, however, proving the contrapositive is much easier (we will show for absolute summability):

$$ \lnot \lr{\sum \limits_{n \in \Z} \abs{h(n)} \lt 0} \Rightarrow \lnot BIBO $$

Note the meaning of a system not being BIBO stable means that $\exists x \in X$ such that $\abs{x(n)} \lt B_x$ for some $B_x \lt \infty$ that produces an output signal that is unbounded. In other words, it produces a $y(n)$ such that $\abs{y(n)} = \infty$ for some $n$.

\begin{proof}

Let $h(n)$ be an impulse response to a system such that $\sum \abs{h(n)} = \infty$. Let $\hat{x}(n)$ be an input signal defined by

$$\hat{x}(n) = \begin{cases} \frac{h(n)}{\abs{h(n)}} & h(n) \neq 0 \\ 0 & \mbox{otherwise} \end{cases}$$

Let $x(n) = \hat{x}(-n)$.

$$x(n) = \begin{cases} \frac{h(-n)}{\abs{h(-n)}} & h(-n) \neq 0 \\ 0 & \mbox{otherwise} \end{cases}$$

Then we have our output signal defined by $y(n) = \sum_{k \in \Z}h(k)x(n-k)$. If we look at $y(0)$, we have
$$y(0) = \sum_{k \in Z}h(k)x(-k)$$
We can use this to show that we will have an unbounded output for any input. 
\begin{align*}
y(0) &= \sum \limits_{k \in Z}h(k)x(-k) \\
y(0) &= \sum \limits_{k \in Z}h(k)\frac{h(k)}{\abs{h(k)}} \\
y(0) &= \sum \limits_{k \in Z}\frac{\lr{h(k)}^2}{\abs{h(k)}} \\
y(0) &= \sum \limits_{k \in Z}\frac{\abs{h(k)}\abs{h(k)}}{\abs{h(k)}} \\
y(0) &= \sum \limits_{k \in Z}\abs{h(k)} = \infty \quad \mbox{(given)}\\
\end{align*}
Therefore, if a system's impulse response is not absolutely summable, the system cannot produce a bounded output when given a bounded input.
\end{proof}

What if $h(n):\Z \to \C$? Then we could have proved the same above by letting $\hat{x}(n) = \frac{h^*(n)}{\abs{h^*(n)}} = \frac{h^*(n)}{\abs{h(n)}}$. This will also "blow up". 

LTI Systems and BIBO Stability