Let $(\Omega, \Sigma)$ be a measurable space. Is the space of bounded measurable functions $B_b(\Sigma)$ equipped with the supremum norm a Banach space, i.e. complete?
Since my comment seems to have caused some confusion, here’s the argument (the answer is yes, of course):
Let $(f_n)_{n \in \mathbb{N}}$ be a Cauchy sequence in $B_{b}(\Omega,\Sigma)$, that is to say:
For every $\varepsilon \gt 0$ there exists $N(\varepsilon)$ such that for all $m,n \geq N$ we have $\|f_n – f_m\|= \sup_{x \in \Omega}{|f_n(x) – f_m(x)|} \lt \varepsilon$.
For each $x \in \Omega$ we have $|f_n(x) – f_m(x)| \leq \|f_n – f_m\|$. It follows that $(f_n(x))_{n \in \mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$. Therefore $f(x) = \lim_{n \to \infty}{f_n(x)}$ exists (i.e. there is a function $f: \Omega \to \mathbb{R}$ such that $f_n \to f$ pointwise).
The pointwise limit of a sequence of measurable functions is measurable: For all $a \in \mathbb{R}$ we have
$$
\begin{align*}
\{x \in \Omega\,:\,f(x) \leq a\}
& =
\left\{
x \in \Omega \,:\,
(\forall k \in \mathbb{N}) \,
(\exists m \in \mathbb{N}) \,
(\forall n \geq m)\quad f_n(x)
\leq a+ \frac{1}{k}
\right\}
\\
& =
\bigcap_{k = 1}^\infty
\bigcup_{m = 1}^\infty
\bigcap_{n = m}^\infty
\left\{x \in \Omega \,:\, f_n(x) \leq a + \frac{1}{k}\right\} \in \Sigma.
\end{align*}
$$
It follows that $f$ is measurable.
For $m,n \geq N(\varepsilon)$ we have for all $x \in \Omega$ that $|f_n(x) – f_m(x)| \lt \varepsilon$. Letting $n \to \infty$ we see that
$$
|f(x) – f_m(x)| \leq \varepsilon \quad\text{ for all } x \in \Omega\text{ whenever } m \geq N(\varepsilon).
$$ It follows that
for all $x \in \Omega$ we have $|f(x)| \leq |f_m(x)| + \varepsilon \leq \|f_m\| + \varepsilon$, so that $f$ is bounded and by point 2 we already know that $f$ is measurable, so $f \in B_b(\Omega,\Sigma)$.
If $m \geq N(\varepsilon)$ we have $\|f – f_m\| \leq \varepsilon$ so that $f$ is the limit of $(f_n)$ in $B_b(\Omega,\Sigma)$.