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Are simple functions dense in $L^\infty$? I've been able to show this for finite measure spaces but not in general.

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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 $\varepsilon$.

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What would happend if $f$ is unbounded? – user62089 Mar 17 '13 at 20:18
@pondy: I'm not sure exactly what you mean to ask. If $f$ is not essentially bounded then the equivalence class of functions a.e. equal to $f$ is not in $L^\infty$. Each element of $L^\infty$ has a bounded representative of its equivalence class. If, say, $f:[0,1]\to\mathbb R$ is an unbounded function, then for every simple function $g:[0,1]\to\mathbb R$, $f-g$ is also unbounded. – Jonas Meyer Mar 17 '13 at 20:39
I am sorry, that was a stupid doubt. Also I wanted to know if all functions in $L^p$ where $p < \infty$ are bounded. Or rather does the implication $\int |f|^p < \infty \Rightarrow |f| < \infty$ hold true. If not is there a counter example. – user62089 Mar 17 '13 at 20:55
@pondy: No. Simple counterexample: $1/\sqrt x$ on $(0,1)$. Basic idea: Function gets big but the sets on which it is big get smaller faster. – Jonas Meyer Mar 17 '13 at 21:08
@pondy: I just remembered the following, related to the question in your last comment:… – Jonas Meyer Mar 17 '13 at 21:23

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 $\parallel f-\phi_n \parallel_{\infty} \rightarrow 0$

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