Inner Product of two functions 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,
Laplace Transform: Continuous analogue of Power series
 A: The $\Delta x$ term is just $1$, as the sum is over the integers. As you shorten the range, it is clear that a $\Delta x$ term is necessary.
A: 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 $n$).
