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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

Brian M. Scott
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Ali
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2 Answers2

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

Brian M. Scott
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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.

Mhenni Benghorbal
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  • The set of all polynomials is not countable. However, the set of polynomials with rational coefficients is, and can be used to approximate any polynomial over $[0, 1]$. – Ayman Hourieh Sep 13 '12 at 09:08
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    I think you should be more explicit about what you mean by "approximated" in the first sentence of the second paragraph (there are two distinct senses used here). Also, the sentence starting with "Hence," could be made much more transparent if you stated what you assume and what you want to conclude. I fear that a novice will be more confused than enlightened by this answer. – t.b. Sep 13 '12 at 09:22
  • @AymanHourieh:Thank you. But as you see I already mentioned "Weierstrass Theorem" which asserts what you said. – Mhenni Benghorbal Sep 13 '12 at 17:56