Are simple functions dense in $L^\infty$? Are simple functions dense in $L^\infty$? I've been able to show this for finite measure spaces but not in general.
 A: If $f$ is bounded, then the function that has value $k\cdot\varepsilon$ on the set where $k\cdot\varepsilon\leq f(x)<(k+1)\cdot\varepsilon$ (for each $k\in\mathbb Z$) is a simple function whose $L^\infty$ distance to $f$ is at most $\varepsilon$.
A: From definition of $L_{\infty}(X,\mathbb{X},\mu)$ as the set of all function essentially bounded , where $f:X\longrightarrow \mathbb{R}$ a function $\mathbb{X}$-measurable is essentially bounded iff there is a bounded function $g:X\longrightarrow \mathbb{R}$ such that $g=f$ on $\mu$-a.e, we have that for any $f\in L_{\infty}$ it can be found a representative $g$ from equivalence class $f$ that is bounded.
So, for that function we can find a sequence $(\phi_n)$ of simple function that converges uniformly to $g$ in $L_{\infty}$, so that sequence converges also to $f$ in $L_{\infty}$. 
As all simple function belongs to $L_{\infty}$, this convergency is given in $L_{\infty}$, so $\parallel f-\phi_n \parallel_{\infty} \rightarrow 0$
A: Consider the function $f \equiv 1$ on $L^{\infty}(\mathbb{R},\mu)$, then $\|s-f\|_{\infty} > \varepsilon$ for any simple function $s$. In particular we can have $\|s-f\|_{\infty} > 1$. So simple functions may not be dense in $L^{\infty}$. But note that here I have assumed that $ \mu \big(\{x \in X : s(x) \neq 0\}\big) < \infty $.
A: By simple approximation theorem(see Real analysis by Royden and Fitzpatrick), if $|f| \leq M$ then there exists a sequence of simple functions $\phi_{\epsilon}, \psi_{\epsilon}$ such that
\begin{eqnarray}
\phi_{\epsilon}(x)\leq f(x) \leq \psi_{\epsilon}(x) \quad \text{for all } x \in \mathbb{R}
\end{eqnarray}
such that $\psi_{\epsilon}(x)-\phi_{\epsilon}(x) \leq \epsilon.$
Now, if $g\in L^{\infty}(\mathbb{R})$ then there exists $\tilde{g}$ such that $g=\tilde{g}$ a.e. and $|\tilde{g}| \leq M$ for some $M>0.$
Thus applying the simple approximation theorem $\phi_{\epsilon}$ converges to $\tilde{g}$ in $L^{\infty}$ which further implies  $\phi_{\epsilon}$ converges to ${g}$ in $L^{\infty}$
