Are simple functions dense in $L^\infty$? I've been able to show this for finite measure spaces but not in general.
Are simple functions dense in $L^\infty$?
2 Answers
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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0@pondy: So on a measure space with finite measure, which simple functions are integrable? On a measure space with infinite measure, can you find a bounded function that is not close to any integrable simple function? – 2013-03-18
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 $\parallel f-\phi_n \parallel_{\infty} \rightarrow 0$