Converting a summation to an integration
You just have to roll through the definitions.  There are some small pitfalls but as long as all limits are defined, which they will be with appropriate adjectives on \$G\$ then we are ok.  There are several definitions of the symbol \$\int\$.  The one which you want to use is the Riemann integral.  To simplify the notation I will set \$F(\omega) = \frac{1}{2\pi}G(\omega)e^{j \omega t}\$.

Now the Riemann integral \$\int_a^b F(\omega) d\omega\$ still has a pretty complicated definition involving sups and infs but  when it does exist and \$a\$ and \$b\$ are integers then it is equal to
$$\int_a^b F(\omega) d\omega =\lim_{j \rightarrow \infty} \sum_{n = aj}^{bj-1} F(n \frac{1}{j})\frac{1}{j}.$$

That is instead of taking \$\Delta \omega \rightarrow 0\$ it is good enough to take \$\Delta \omega = \frac{1}{j} \$ with \$j \rightarrow \infty \$. This is the definition you learned in Calculus class.

The improper integral is defined as 
$$\int_{-\infty}^{\infty}F(\omega) d\omega = \lim_{a \rightarrow -\infty , b \rightarrow \infty} \int_a^b F(\omega)d\omega.$$

Again, if this limit exists then this is just the limit (for \$k\$ an integer)
$$\int_{-\infty}^{\infty}F(\omega)d\omega = \lim_{k \rightarrow \infty}\int_{-k}^k F(\omega)d\omega.$$

Combining all of this and not discussing when/how to changing the order of taking limits is valid we obtain
$$\begin{eqnarray*}
\int_{-\infty}^{\infty} F(\omega) d\omega &amp;=&amp; \lim_{k \rightarrow \infty} \int_{-k}^k F(\omega)d\omega\\
&amp;=&amp; \lim_{k \rightarrow \infty}\lim_{j \rightarrow \infty} \sum_{n = -kj}^{kj-1} F(n \frac{1}{j})\frac{1}{j}\\
&amp;=&amp; \lim_{j \rightarrow \infty}\lim_{k \rightarrow \infty} \sum_{n = -kj}^{kj-1} F(n \frac{1}{j})\frac{1}{j}\\
&amp;=&amp; \lim_{j \rightarrow \infty} \sum_{n = -\infty}^{\infty} F(n \frac{1}{j})\frac{1}{j}\\
&amp;=&amp; \lim_{\Delta \omega \rightarrow 0} \sum_{n = -\infty}^{\infty} F(n \Delta \omega)\Delta \omega
\end{eqnarray*}$$