Using characteristic functions to determine distribution of sum of independent normal random variables.

There is a bijective correspondence between characteristic functions and probability distributions. It is stated in Probability Theory by Durrett that from this fact it follows readily that given independent normal random variables $X_1$, $X_2$ with mean $0$ and variances $sigma_1^2$ and $sigma_2^2$ respectively, the sum $X_1+X_2$ is a normal random variable with mean $0$ and variance $sigma_1^2+sigma_2^2$. I imagine this proof is something like:

The ch. f. of $X_1$ is $$phi_1(t)= text{exp}Big(frac{-sigma_1^2 t^2}{2}Big).$$

The ch. f. of $X_2$ is $$phi_2(t)= text{exp}Big(frac{-sigma_2^2 t^2}{2}Big).$$

The ch. f. of $X_1+X_2$ is $$phi_1(t)phi_2(t)= text{exp}Big(frac{-sigma_1^2 t^2}{2}Big)text{exp}Big(frac{-sigma_2^2 t^2}{2}Big)=text{exp}Big(frac{-(sigma_1+sigma_2^2) t^2}{2}Big),$$

which is the ch.f. of a normal random variable with variance
$sigma_1^2+sigma_2^2$, and thus, as a ch. f. determines a
unique distribution, $X_1+X_2$ must be a normal random variable with variance
$sigma_1^2+sigma_2^2$.

The only thing that troubles me is that it seems strange to me that the exact same proof seems to go through without the assumption of $EX_1=EX_2=0$. Taking $mu_1=EX_1$ and $mu_2=EX_2$ we get that:

The ch. f. of $X_1$ is $$phi_1(t)= text{exp}Big(frac{-sigma_1^2
t^2}{2}+imu_1tBig).$$

The ch. f. of $X_2$ is $$phi_2(t)= text{exp}Big(frac{-sigma_2^2
t^2}{2}+imu_2tBig).$$

The ch. f. of $X_1+X_2$ is $$phi_1(t)phi_2(t)=
text{exp}Big(frac{-(sigma_1+sigma_2^2)
t^2}{2}+(mu_1+mu_2)itBig),$$

which is the ch. f. of a normal random variable with variance
$sigma_1^2+sigma_2^2$ and mean $mu_1+mu_2$, and thus, as a ch. f.
determines a unique distribution, $X_1+X_2$ must be a normal random variable
with variance $sigma_1^2+sigma_2^2$ and mean $mu+mu_2$.

Is there something wrong with this second argument?