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In fact I am reading the book of Ohsawa, Analysis of Several Complex Variables, and I came across this line on page 13,

... $L^{2}_\mathrm{loc}(\Omega)$ with respect to the topology induced by the $L^2$ convergence on compact sets.

Yet I am not clear how to induce a topology from. I have a guess. In the following I am considering the case $\Omega=\mathbb{R}$. Define $||f||_n=\int_{-n}^n |f|$ for $f\in L^2_\mathrm{loc}(\mathbb{R}), n=1,2,3,\ldots$. Put $||f||=\sum\limits _{n=1}^\infty\frac{||f||_n}{1+||f||_n} \frac{1}{2^n}.$ Then $||\cdot||$ is a metric for $L_\mathrm{loc}^2 (\mathbb{R}) $. Is it this topology? Or something else? Would someone be kind enough to give me some hints on this? Thank you very much.

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    A keyword to search for is «final topology». The general nonsense attached to that is extremely useful when dealing with this situation and —in my experience— intuitively natural.2012-01-10

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While it is nice to have a metric, it's often not very useful. For example if $\Omega \subseteq \mathbb R^d$ is open, $C^\infty(\Omega)$ can be given a metric that turns it into a Fréchet space, but it's much simpler (and often more useful) to state what the mode of convergence on the space is: uniform convergence of $f_n$ and all its derivatives on compact subsets of $\Omega$.

In the language of topological vector spaces, we can say that we're specifying the topology on $C^\infty(\Omega)$ as the initial topology with respect to the seminorms $\lVert f\rVert_{K,\alpha} := \sup_{x \in K}|D^\alpha f(x)|$ , where $K$ ranges over the compact subsets of $\Omega$ and $\alpha$ ranges over multiindices in $\mathbb N_0^d$.

Contrast this with how hard it is to work with a metric on the space: first we have to find a countable family of nested compact sets $(K_n)$ that exhaust $\Omega$, which there is no canonical way of doing. Then we have to define our metric as (something like) $d(f,g) := \sum_{n=1}^\infty 2^{-n}\min\{\sup\limits_{x\in K_n} \max\limits_{|\alpha| \leq n} \left|D^\alpha (f-g)\right|,1\},$

and I think we can all see that proving that the topology produced by this metric is independent of the choice of the $K_n$ is rather irksome.

In other words, while your construction does indeed work, it's kind of missing the point: Just saying that $L^2_{\mathrm{loc}}$ has the topology of $L^2$-convergence on compact subsets of $\mathbb R^n$ is enough to simultaneously specify the topology and describe its most useful feature.