Show that this is an inner product Let's define
$$(f,g)=\int_{\mathbb{R}} \frac{f(x)\bar{g}(x)}{1+x^2}dx$$
$\forall f,g\in X=\{h:\mathbb{R}\rightarrow\mathbb{C}:$ $h$ is Lebesgue-measurable and bounded over $\mathbb{R}$}
I have to show that $(\cdot,\cdot)$ is an inner product over $X$ (I've already shown that $X$ is a vector space over $\mathbb{C}$).
It is obvious that, $\forall f,g,h\in X$, $(f+g,h)=(f,h)+(g,h)$ and, $\forall \lambda\in\mathbb{C}$, that $(\lambda f,g)=\lambda(f,g)$. 
It's also clear that $(f,g)=\overline{(g,f)}$, so we obtain the skew-linearity in the second argument. Now I should show that $(f,f)\geq 0$ $\forall f\in X$ (this is trivial) and that $(f,f)=0 \Rightarrow f=0$, but this seems to be false, because any $f\in X$ which is almost everywhere $0$ over $\mathbb{R}$ would be such that $(f,f)=0$, even if $f$ is not identically $0$. Did I do some mistakes in this argument?
 A: You are right that $(f, f)=0$ doesn't implies that $f\equiv 0$ as a function. Nevertheless, there's a sense in which $(\cdot, \cdot)$ can be an inner product in some space.
Consider the space $\mathcal L^2\left(\Bbb R,\frac 1{1+x^2}\,dx\right)$ consisting of functions $f:\Bbb R\to \Bbb C$ such that
$$
\int_{\Bbb R} \frac {|f|^2}{1+x^2}dx < \infty.
$$
We can then quotient out by the relation $g\sim f$ iff $f=g$ almost everywhere. Then the elements of
$$
L^2\left(\Bbb R,\frac 1{1+x^2}\,dx\right):= \mathcal L^2\left(\Bbb R,\frac 1{1+x^2}\,dx\right)\Big/ \sim
$$
would consist of equivalent classes of functions instead.
Now, for $f\in L^2\left(\Bbb R,\frac 1{1+x^2}\,dx\right)$ if $(f, f)=0$ then, as you've shown, we would have $f=0$ almost everywhere. This means that $f$ belong to the equivalent class of the constant function $0$, so $f$ is the zero vector in $ L^2\left(\Bbb R,\frac 1{1+x^2}\,dx\right)$.
Note that this process of quotient out by $\sim$ as I mentioned is the usual one used in the construction of $L^p(\Omega)$ spaces in functional analysis.
