The inner product of two vectors $\vec a$ and $\vec b$ of n dimensions, is given by, $$(\vec a \,, \vec{b}) =a_1b_1 + a_2b_2 + a_3b_3 + \,\,...\,\,+a_nb_n $$
If a function is considered to be a vector of infinite dimensions, for example, $$f(x) = \begin{bmatrix} \vdots \\ f(0.05) \\ f(0.051) \\ f(0.052) \\ \vdots \end{bmatrix}$$ Then, the inner product of two functions say $f(x), g(x)$ is given by $$\mathbf {\bigl(} f(x),g(x)\mathbf {\bigr)}=\int_{-\infty}^\infty f(x)g(x) \,\,\, \mathbf{dx}$$ My Question is:
How did we get the $\mathbf {dx}$ in the inner product (or)$$ \mathbf{why\,\, is}\,\,\,\,\ \sum_{-\infty}^\infty f(x)g(x) = \int_{-\infty}^\infty f(x)g(x) \,\,\, \mathbf{dx}$$ In a Riemann sum , the $\Delta x $ term becomes $dx$ during the limiting process, but in this sum, there doesn't seem to be a $\Delta x$ term. I think the answer to this question would probably answer my another question,
We use choices of inner product that (a) indeed satisfy the axioms defining an inner product and (b) prove useful for our purposes. Under appropriate regularisation conditions, $\sum_{n=-\infty}^\infty f(n)g(n)$ (the sum implied to run over integers only, since you can't sum uncountably many terms) and $\int_{-\infty}^\infty f(x)g(x)dx$ are both inner products. But the latter is of greater interest when studying $\Bbb R\mapsto\Bbb R$ functions, since it runs over all values the arguments of $f,\,g$ can take.
In particular, (c) this integral is an inner product (proof is an exercise) and (d) you can't have an integral without the $dx$ part. What you can do, however, is replace $dx$ with the more general $dh(x)=h^\prime(x) dx$ to obtain $\int_{-\infty}^\infty f(x)h^\prime(x)g(x) dx$, which is an inner product provided $h^\prime(x)>0$ for all $x\in\Bbb R$. (Similarly, $\sum_n f(n)j(n)g(n)$ is an inner product if $j(n)>0$ for $n\in\Bbb Z$).
So, "why $dx$?" really means "why take $h(x)=x$ for all $x$?" To which we can only answer, "use an $h$ that's convenient". For example, you'd use a very different $h$ if studying these, whereas in studying these you'd take $h(x)=x$ but you'd restrict the integration range $[-1,\,1]$ (which still gives an inner product, just as in the discrete case one can restrict to summing over fewer $n$).