Topology of $p$-integrable functions space. In a reference, I read that topology of $L^p(\mathbb{R}^n)$, $1\leq p\leq \infty$. What is the topology of $L^p(\mathbb{R}^n)$?
I know that $f\in L^p$, $\|f\|_{p}^{p}=\int_{\mathbb{R}^n}|f(x)|^{p}dx$. Then, the topology is $\tau=\left\{B(f,r):f\in L^p, r>0\right\}$?
 A: Generally speaking, any metric space $(X,d)$ has a topological structure. The natural topology (associated to the distance $d$) is the set of all 'open' sets for distance $d$, i.e. the set
$$
\mathcal U = \{ A \subset X : \forall y \in A, \ \exists \varepsilon > 0, \ B(y,\varepsilon) \subset A\}
$$
where $B(y, \varepsilon) = \{ x \in X : d(x,y) < \varepsilon\}$.
You can quite easily show that $\mathcal U$ is indeed a topology on $X$.
Of course, this applies to normed vector spaces, since to any norm $\| \cdot \|$ you can associate the natural distance $d : (x,y) \mapsto \|x-y\|$.
So in your case, you can immediately see how the natural topology is defined on $L^p(\mathbb R^n)$ equipped with the $L^p$ norm.
Note that the topology associated to a normed vector space WILL depend on the norm you choose on that space. However, it can be shown that the norm you choose will not matter if you vector space is of finite dimension (this is a direct consequence of the fact that all norms are equivalent in a finite dimension vector space). Nonetheless, $L^p(\mathbb R^n)$ is obviously not a finite dimension vector space so this does not hold in your case.
A: I hope this helps to clarify some matters:

*

*On $L^p(\mathbb{R}^n)$, $\|\;\|_p$ defines a pseudonorm, that is, a  $\|\;\|_p$ is a nonnegative, homogeneous, and satisfies the triangle inequlity. However, $\|f-g\|_p=0$ does does not mean that $f(x)=g(x)$ for all $x\in\mathbb{R}^n$. A typical example in $\mathbb{R}^1$ is $f(x)=\mathbb{1}_{\mathbb{Q}}(x)$ for which $\|f\|_1=0$.


*As a consequence, the topology generated by the pseudonym $\|\;\|_q$ (for which the  collection of balls $B(f;r)$ with $f\in L^p(\mathbb{R}^n)$, $r>0$, is a basis) is not adequate for convergence for it would not be a $T_1$ topology (and thus Hausdorff either).


*Thus, it is  convenient to first identify functions $f$ and $g$ if $\|f-g\|_p=0$. Denote this (identified) space as $L_p(\mathbb{R}^n)$ to differentiate it with the original one (I just change superscript by subscript)


*On $L_p(\mathbb{R}^n)$, $\|\;\|_p$ is a  norm (also complete norm by the way, but that is not very relevant or your question). So the topology there is that generated by the metric defined by the norm, i.e.
$$d(f,g):=\|f-g\|_p$$

*

*A base for the topology is the one stated in your question, but with $L_p$ instead of $L^p$.

