Going back to the basics for a second. Say we have a random variable Y that is normally distributed: $X$~$N(\mu_1,\sigma_1^2)$. That means the pdf of X is: $$f_X(x)=\frac{1}{\sqrt{2\pi}\sigma_1}e^{-\frac{1}{2}(\frac{x-\mu_1}{\sigma_1})^2}$$ We also have a variable Y that is normally distributed: $Y$~$N(\mu_2,\sigma_2^2)$, which means the pdf of Y is: $$f_Y(y)=\frac{1}{\sqrt{2\pi}\sigma_2}e^{-\frac{1}{2}(\frac{y-\mu_2}{\sigma_2})^2}$$
We're told that if X and Y are independent, then for any set of constants $a_1$ and $a_2$, $Z=a_1X+a_2Y$ is normally distributed with mean and variance given by: $Z$~$N(\mu_1+\mu_2,a_1^2\sigma_1^2+a_2^2\sigma_2^2)$
Let's take the case when $a_1$=$\sqrt{2\pi}\sigma_1$ and $a_2=0$. Then $Z=\sqrt{2\pi}\sigma_1X$ and its pdf is: $$f_Z(x)=\sqrt{2\pi}\sigma_1 \frac{1}{\sqrt{2\pi}\sigma_1}e^{-\frac{1}{2}(\frac{x-\mu_1}{\sigma_1})^2}=e^{-\frac{1}{2}(\frac{x-\mu_1}{\sigma_1})^2}$$
If $f_X(x)$ integrates to 1, then $f_Z(x)$ doesn't. We say a variable Z is normally distributed with mean $\mu$ and variance $\sigma^2$ if its pdf is given by $$f_Z(z)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2}(\frac{z-\mu}{\sigma})^2}$$ Well, we've shown above that it isn't. So why do we say Z is normally distributed?
