Let $\mu$ be a complex measure on $\Bbb{T}$ and $f:\Bbb{D}\to\Bbb{C}$ be the harmonic function defined as poisson integral of $\mu$ i.e. $$f_r(\theta)=f(re^{i\theta}):=\int P_r(\theta-t)\ d\mu(t)$$ where $P_r$ denotes the Poisson Kernel defined by the formula $P_r(\theta)=\text{Re}\left(\frac{1+re^{i\theta}}{1-re^{i\theta}}\right)$
We have to prove that $\lim\limits_{r\to1}\lVert f_r\rVert_1=\lVert\mu\rVert$
$\lVert f_r\rVert_1=\frac{1}{2\pi}\int\limits_{-\pi}^{\pi} |f_r(\theta)|\ d\theta$
$=\frac{1}{2\pi}\int\limits_{-\pi}^\pi \left|\int\limits_{-\pi}^\pi P_r(\theta-t)\ d\mu(t)\right|\ d\theta$
$\le \frac{1}{2\pi}\int\limits_{-\pi}^\pi \int\limits_{-\pi}^\pi P_r(\theta-t)\ d\theta\ d|\mu|(t)$
$=\int\limits_{\pi}^\pi d|\mu|(t)$
$=|\mu|(\Bbb{T})=\lVert \mu\rVert$
This implies $\lim\limits_{r\to 1}\lVert f_r\rVert_1\le\lVert\mu\rVert$
Now I want to prove the converse inequality. I know that from Fatou's theorem
$$\lim\limits_{r\to1} f_r(\theta)=\tilde{f}(\theta)\ \text{for almost every }\theta$$ where $\tilde{f}\in L^1(\Bbb{T})$ is the derivative of $\mu$ with respect to the normalized Lebesgue measure i.e. $d\mu=\frac{1}{2\pi}\tilde{f}\ d\theta+d\mu_s$ where $\mu_s$ is singular.
I think from this, somehow I need to establish the other inequality. Can anyone help me with an idea to finish the proof? Thanks for your help in advance.
Let $\psi \in C^0(\mathbb{T})$. Suppose we can show that $\int_{\mathbb{T}}{f_r(\theta)\psi(\theta)d\theta} \rightarrow 2\pi\int_{\mathbb{T}}{\psi d\mu}$. Then, as continuous linear forms on $C^0(\mathbb{T})$, $\frac{1}{2\pi}f_r(\theta) d\theta \rightarrow d\mu$, so for the operator norm, $\|d\mu\| \leq \liminf_{r \rightarrow 1} \|\frac{1}{2\pi}f_r(\theta)d\theta\|$, which is the inequaily we need, ie $\|\mu\| \leq \liminf_{r \rightarrow 1} \|f_r\|_1$.
The convergence statement that we need is linear wrt $\psi$ and linear wrt $\mu$, so it’s enough to prove it when $\psi$ and $\mu$ are nonnegative.
Then $$\int_{\mathbb{T}}{f_r(\theta)\psi(\theta)d\theta}=\int_{\mathbb{T}^2}{P_r(x-y)\psi(x)dxd\mu(y)}=\int_{\mathbb{T}}{\int_{\mathbb{T}}{P_r(x)\psi(x+y)dx}d\mu(y)}.$$
For all $y \in \mathbb{T}$ and $0<r <1$, as $P_r$ is non-negative, $\left|\int_{\mathbb{T}}{P_r(x)\psi(x+y)dx}\right| \leq \|\psi\|_{\infty}\int_{\mathbb{T}}{P_r(x)dx} \leq 2\pi\|\psi\|_{\infty}$.
So, by dominated convergence, we just need to show that for almost all $y$, $\int_{\mathbb{T}}{P_r(x)\psi(x+y)dx} \rightarrow 2\pi\psi(y)$.
But this integral is $\int_{\mathbb{T}}{P_r(x-y)\psi(x)dx}=\int_{\mathbb{T}}{P_r(y-x)\psi(x)dx}=g(re^{iy})$, where $g$ is the Poisson integral of $\psi(\theta)d\theta$. The conclusion then follows from Fatou’s theorem, as in your post (we don’t actually need it, there’s an elementary argument showing that convergence holds for all $y$).