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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
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@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: math.stackexchange.com/questions/90668/… –  Jonas Meyer Mar 17 '13 at 21:23
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