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I have no idea how to prove that the space $X$ of all integrable functions on the interval $[0,1]$, for $f,g\in X$, with the following metric:

$\rho(f,g)=\int|f(x)−g(x)|~dx$

is separable. I'll appreciate any help, please. Thank you

2 Answers 2

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HINT: Look at step functions whose steps occur at rationals in $[0,1]$ and which take on only rational values.

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A topological space is separable if it contains a countable dense subset.

You can use the fact that:

integrable functions on $[0,1]$ can be approximated by continuous functions and any continuous function on $[0,1]$ can be approximated by a polynomial with rational coefficients (Weierstrass Theorem). Noting that, the collection of all such polynomials is dense in $C[0,1]$ and it is a countable set.

Hence, we have

$ \rho(f,p)=\int|(f(x)-c(x))+(c(x)-p(x))|~dx $ $\leq \int |f(x)-c(x)|dx + \int|c(x)-p(x)|~dx $

$\leq \int |f(x)-c(x)|dx + \int \sup|c(x)-p(x)|~dx < \frac{\epsilon}{2} + \frac{\epsilon}{2}=\epsilon\,, $

where $p(x)$ is a polynomial with rational coefficients and $c(x) \in C[0,1]$. This implies that the space of integrable functions is separable.

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    @AymanHourieh:Thank you. But as you see I already mentioned "Weierstrass Theorem" which asserts what you said.2012-09-13