= Solution
The <Mazur theorem> states that the <weak closure> and norm closure of a <convex set> $C$ in a <normed vector space> coincide:
$$
\boxed{\overline C^{\,w}=\overline C^{\,\|\cdot\|}.}
$$
Norm closure is contained in <weak closure> because the <weak topology> is coarser. Conversely, if $x\notin\overline C^{\,\|\cdot\|}$, the <Hahn-Banach separation theorem> provides a continuous real linear functional strictly separating $x$ from that closed <convex set>. In a complex space this is the <real part> of a continuous complex linear functional. A weak neighborhood of $x$ then misses $C$, so $x\notin\overline C^{\,w}$. This proves the equality. It also gives the usual <Mazur lemma>: if $x_n\rightharpoonup x$, then $x$ lies in the norm closure of the <convex hull> of each tail, so one can choose tail <convex combinations> $y_n$ with $\|y_n-x\|<1/n$.
Now let $K$ be a <weakly compact set> in a normed space $X$. Each $\varphi\in X^*$ is bounded on $K$ because it is weakly continuous. The family $\{Jx:x\in K\}$ in $\mathcal B(X^*,\mathbb F)$, where $J$ is the <canonical embedding into the bidual>, is therefore pointwise bounded. The space $X^*$ is Banach even if $X$ is not. Apply the <Uniform boundedness principle> and use $\|Jx\|=\|x\|$ to obtain
$$
\boxed{\sup_{x\in K}\|x\|<\infty.}
$$
This proves that a <weakly compact set is norm bounded> without assuming completeness of the original space.
For the real-valued dual and integral formulas that follow, take $X$ to be real, as in the PDF. In a complex space the norming formula uses <real parts>, and the integral identities use complex-valued functionals instead.
If the <separable Banach space> $X$ is nonzero, choose a norm-dense sequence $(u_n)$ in its unit sphere. By the <Hahn-Banach theorem>, choose $\varphi_n\in X^*$ with $\|\varphi_n\|=1$ and $\varphi_n(u_n)=1$. For any unit vector $u$, arbitrarily close $u_n$ satisfy
$$
1-\|u-u_n\|\leq\varphi_n(u)\leq1.
$$
Scaling gives the <countable norming family> identity
$$
\boxed{\|x\|=\sup_n\varphi_n(x)\quad(x\in X).}
$$
For $X=\{0\}$ use the constant sequence of zero functionals. If $f:\Omega\to X$ is norm-Borel measurable, every $\varphi_n\circ f$ is measurable, so its countable supremum $\|f(\cdot)\|$ is measurable. Equivalently, this also follows directly from continuity of the <norm>.
For any $\varphi\in X^*$, continuity makes $\varphi\circ f$ measurable, and
$$
|\varphi(f(\omega))|\leq\|\varphi\|\,\|f(\omega)\|.
$$
Thus the assumed integrability of the norm implies scalar integrability, and
$$
T_f(\varphi)=\int_\Omega\varphi(f(\omega))\,d\mu(\omega),\qquad
|T_f(\varphi)|\leq\|\varphi\|\int_\Omega\|f\|\,d\mu
$$
defines a bounded linear functional on $X^*$. Use the granted weak-star continuity of $T_f$. By the <continuous dual of a weak-star topology>, $T_f$ is evaluation at a vector of $X$. Indeed, continuity gives finitely many $x_1,\ldots,x_m$ and $\varepsilon>0$ such that $|T_f(\varphi)|<1$ whenever $|\varphi(x_j)|<\varepsilon$ for all $j$. Scaling shows that $T_f$ vanishes on the common kernel of these evaluations. It therefore factors through their finite-dimensional coordinate map, so $T_f(\varphi)=\sum_ja_j\varphi(x_j)=\varphi(\sum_ja_jx_j)$. The <Hahn-Banach theorem> makes this representing vector unique. Hence
$$
\boxed{v=\int_\Omega f\,d\mu\in X,\qquad \varphi(v)=\int_\Omega\varphi\circ f\,d\mu\quad(\varphi\in X^*),\qquad
\|v\|\leq\int_\Omega\|f\|\,d\mu.}
$$
In this separable setting the vector is the <Bochner integral>.
Return to a <weakly compact set> $K\subseteq X$ and its inclusion $f:K\to X$. For each fixed $a\in X$, the identity $\|x-a\|=\sup_n\varphi_n(x-a)$ makes $x\mapsto\|x-a\|$ weakly Borel measurable. Norm balls are consequently weakly Borel measurable. Separability gives a countable base of such balls, so every norm-open set is weakly Borel measurable. This proves measurability of $f$, and establishes the equality of the <weak and norm Borel sigma-algebras in a separable Banach space>.
Put $M=\sup_{x\in K}\|x\|<\infty$. For every finite signed <Borel measure> $\mu$ on $K$,
$$
\boxed{\int_K\|f(x)\|\,d|\mu|(x)\leq M|\mu|(K)<\infty.}
$$
For positive measures this is the integral in the question. For signed measures, the correct integrability condition uses the <variation measure>; define the integral by taking the difference of the positive and negative integrals. The printed $\Omega$ in this clause should be $K$, the domain of the inclusion.
The <Riesz-Markov-Kakutani representation theorem> now defines the bounded linear map
$$
T:C(K)^*\to X,\qquad T(\mu)=\int_Kx\,d\mu(x),\qquad \|T(\mu)\|\leq M\|\mu\|_{\mathrm{TV}}.
$$
For each $\varphi\in X^*$ the restriction $\varphi|_K$ is in $C(K)$, and
$$
\varphi(T(\mu))=\mu(\varphi|_K).
$$
The right side is weak-star continuous in $\mu$. The defining property of the <weak topology> therefore proves that $T$ is weak-star-to-weak continuous, for arbitrary nets. For a <Dirac measure>, $\boxed{T(\delta_x)=x}$.
If $K\ne\varnothing$, let $\mathcal P(K)$ be its regular <probability measures>. This is a weak-star closed subset of $B_{C(K)^*}$: its conditions are $\mu(1)=1$ and $\mu(g)\geq0$ for every nonnegative $g\in C(K)$. It is compact by <Banach-Alaoglu theorem>. Thus $T(\mathcal P(K))$ is weakly compact and <convex>, and contains $K$ because it contains all $T(\delta_x)$. It is weakly closed, hence norm closed, so it contains $C=\overline{\operatorname{conv}K}^{\,\|\cdot\|}$. By <Mazur theorem>, $C$ is weakly closed. Therefore it is a closed subset of the weakly compact set $T(\mathcal P(K))$, proving
$$
\boxed{\overline{\operatorname{conv}K}^{\,\|\cdot\|}\text{ is weakly compact}.}
$$
The empty case is immediate. In fact $T(\mathcal P(K))=C$: a <barycenter of a measure on a Banach space> outside $C$ would be strictly separated by a functional $\varphi$, contradicting $\varphi(T(\mu))=\int_K\varphi\,d\mu\leq\sup_K\varphi$.
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