Proof that a probability distribution is uniquely determined by its characteristic function.

Question

I'm currently working through the proof in my lecture notes for the statement that a probability distribution is uniquely determined by its characteristic function. However, I got stuck on quite a few points.

One lemma necesarry for this proof used states the following:

Proposition $10.12$ Let $P$ be a probability distribution on $\mathbb{R}^{d} .$ Let $P^{\sigma}=P * \mathcal{N}\left(0, \sigma^{2} I\right)=$ $\mathcal{L}(X+\sigma Z)$ where $X, Z$ are independent with $\mathcal{L}(X)=P$ and $\mathcal{L}(Z)=\mathcal{N}\left(0, \sigma^{2} I\right), \sigma>0$ Then $\hat{P}^{\sigma}(t)=\hat{P}(t) e^{-\frac{\|t\|^{2} \sigma^{2}}{2}}$ and $P^{\sigma}$ has the density $$ f^{\sigma}(x)=\frac{1}{(2 \pi)^{d}} \int_{\mathbb{R}^{d}} \hat{P}(-t) e^{-\frac{\|t\|^{2} \sigma^{2}}{2}} e^{i<x, t>} d t . $$

However there actually seems to be some notation error in the lectures notes because later on in the proof $\mathcal{L}(Z)$ is said to be $\mathcal{N}\left(0,I\right)$.

Theorem $10.13$ Let $P$ be a probability distribution on $\mathbb{R}^{d} .$ Then it is uniquely determined by $\hat{P}$. Moreover

1. for each bounded and $P$ -a.s. continuous function $h$ we get $$ \int_{\mathbb{R}^{d}} h(x) d P(x)=\lim _{\sigma \rightarrow 0} \int f^{\sigma}(x) h(x) d x $$

Proof.

1. Clearly $\int f^{\sigma}(x) h(x) d x=E(h(X+\sigma Z))$ where $X$ and $Z$ are independent with $\mathcal{L}(X)=$ $P$ and $\mathcal{L}(Z)=\mathcal{N}(0, I) .$ By dominated convergence this converges to $E(h(X))=$ $\int h(x) d P(x)$ as $\sigma \rightarrow 0$. Uniqueness follows, since $\hat{P}$ determines all $f^{\sigma}$ and hence all integrals $\int h d P$.

Now my questions regarding the proof are the following:

1. Why is $f^{\sigma}(x) h(x) d x=E(h(X+\sigma Z))$ and not $E(h(X+Z))$?

2. Why is $E(h(x))=\int h(x) d P(x)$? Specifically, why are we integrating over $P$ all of a sudded instead of $x$?

If my interpretation is correct, this part of the proof (there is a second statement regarding distributions with densities) shows that characteristic functions uniquely determine their corresponding probability distributions because we can set $h(x) = 1_{(-\infty,x]}$ and get the cumulative distribution function back.