Why are bounded functions dense? My aim is  to find the closure of  $l^p$ is  in $l^q$  ( where $p< q$) .It seems to me we  know $l^q$ is seperable so there is a  such subset M of $l^q$  such that closure of $M=l^p$. I want to show $M=l^p$ .Is it possible to do it ? 
But  Woodface  said 'It's all of $l^q$. Prove that bounded functions are dense'.
 A: A few corrections:


*

*On a finite measure space $L^p$ is dense in $L^q$ when $p > q$. So you have this inequality backwards.

*On discrete measure spaces, the hierarchy goes the other way, and $L^p$ is dense in $L^q$ when $p<q$. When the space is infinite, this hierarchy is strict. $\mathbb{N}$ with the counting measure is the canonical example here, to the point that $L^p$ in this setting is commonly denoted by $\ell^p$. This also generalizes directly to atomic measure spaces, where all positive measure subsets have at least some specified measure.

*On an infinite measure space which is not purely atomic, there is no hierarchy of $L^p$ spaces. This is because large $p$ amplifies singularities but dampens out infinite tails. $\mathbb{R}$ with the Lebesgue measure is the canonical example of such a space. The two example functions to keep in mind are as follows. Take $c \in (0,1)$. Then $x^{-c} \chi_{[0,1]}$ is in $L^p$ only for small $p$ while $x^{-c} \chi_{[1,\infty)}$ is in $L^p$ only for large $p$. 


As for your original question, there are a number of views. One "low level" view notes that simple functions are dense and bounded. This "simple approximation theorem" is either one of the foundational results proven early on in the theory of Lebesgue integration, or is a definition. I call this view "low level" because the way you discuss it depends on the specifics of your definitions of Lebesgue integration.
A "high level" view is to use the Lebesgue dominated convergence theorem to build a concrete approximating sequence. For instance, $f_n = \text{sign}(f) \cdot \min \{ |f|,n \}$ is a sequence of bounded functions which converges to $f$. I call this view "high level" because it depends only on the theorems of Lebesgue integration, not on the specifics of the construction.
A: Basically the reason is that given an unbounded function $f$ in $L^p$, you can make it bounded by modifying it on $\{x: |f(x)| > N\}$ (say making it $0$ there), and if $N$ is large the $L^p$ distance from $f$ to the modified function is  small.
