Consider $X(j\omega)\$ evaluated according to Equation 4.9:

$X(j\omega) = \int_{-\infty}^\infty x(t)e^{-j \omega t} dt$

and let $x(t)\$ denote the signal obtained by using $X(j\omega)\$ in the right hand side of Equation 4.8:

$x(t) = (1/(2\pi)) \int_{-\infty}^\infty X(j\omega)e^{j \omega t} d\omega$

If $x(t)\$ has finite energy, i.e., if it is square integrable so that Equation 4.11 holds:

$\int_{-\infty}^\infty |x(t)|^2 dt < \infty$

then it is guaranteed that $X(j\omega)\$ is finite, i.e, Equation 4.9 converges.

Let $e(t)\$ denote the error between $\hat{x}(t)\$ and $x(t)\$, i.e. $e(t)=\hat{x}(t) - x(t)\$, then Equation 4.12 follows:

$\int_{-\infty}^\infty |e(t)|^2 dt = 0$

Thus if $x(t)\$ has finite energy, then although $x(t)\$ and $\hat{x}(t)\$ may differ significantly at individual values of $t\$, there is no energy in their difference.