On the convexity of a particular form of integrals EDIT: I made some critical corrections below.
Let $\mathcal{H}\colon\mathbf{w}\cdot\mathbf{x}+c=0$ be a hypeplane in $\mathbb{R}^n$. Also, let $g\colon\mathbb{R}^n\to\mathbb{R}_+$, be a non-negative, real-valued function. I would like to decide on the convexity of the function $f\colon\mathbb{R}^n\times\mathbb{R}\to\mathbb{R}$, given by
$$
f(\mathbf{a},b)=\int_{\Omega}\! (\mathbf{a}\cdot\mathbf{x}+b)g(\mathbf{x}) \,\mathrm{d}\mathbf{x},
$$
where $\mathbf{x}\in\mathbb{R}^n$, $b\in\mathbb{R}$, and $\Omega$ is the half-space defined by $\mathcal{H}$ as $\Omega=\{\mathbf{x}\in\mathbb{R}^n\rvert\mathbf{w}\cdot\mathbf{x}+c\geq0\}$.
What I have thought so far is to express the integral as a sum (if such a thing is feasible) and use the property of the affine (and consequently convex) quantity $(\mathbf{a}\cdot\mathbf{x}+b)\cdot k$, where $k\in\mathbb{R}$. Albeit, I am not sure at all this is going to work. In addition, I am not sure in what direction should I work to...
Is there any appropriate argument I could use instead?

Let me make my question more general and clear, maybe giving an answer, thanks to @p.s, of course!
Let $f\colon K\subseteq\mathbb{R}^n\times\mathbb{R}\to\mathbb{R}$, given by
$$
f(\mathbf{a},b)=\int_{\Omega}\! h(\mathbf{a},b,\mathbf{x})g(\mathbf{x})  \,\mathrm{d}\mathbf{x},
$$
where $\Omega=\{\mathbf{x}\in\mathbb{R}^n\rvert\mathbf{w}\cdot\mathbf{x}+c\geq0\}$. Also, let $g\colon\mathbb{R}^n\to\mathbb{R}_+$ be a non-negative function.
If $h$ is convex with respect to $\mathbf{a}$, $b$ over the domain $\Omega$, then $f$ is convex in $K$.
Proof
Let $\lambda\in[0,1]$ and $(\mathbf{a}_1,b_1),(\mathbf{a}_2,b_2)\in K$. Then,
$$
\begin{aligned}
f(\lambda\mathbf{a}_1+(1-\lambda)\mathbf{a}_2,\lambda b_1+(1-\lambda)b_2) &= \int_\Omega\!h(\lambda\mathbf{a}_1+(1-\lambda)\mathbf{a}_2,\lambda b_1+(1-\lambda)b_2,\mathbf{x})g(\mathbf{x})\,\mathrm{d}\mathbf{x} \\
                           &\le \int_\Omega\!\left[\lambda h(\mathbf{a}_1,b_1,\mathbf{x}) + (1-\lambda) h(\mathbf{a}_2,b_2,\mathbf{x})\right]g(\mathbf{x})\,\mathrm{d}\mathbf{x} \\
                           &= \lambda \int_\Omega h(\mathbf{a}_1,b_1,\mathbf{x}) g(\mathbf{x})d\mathbf{x} + (1-\lambda) \int_\Omega\!h(\mathbf{a}_2,b_2,\mathbf{x})g(\mathbf{x})\,\mathrm{d}\mathbf{x} \\
                           &= \lambda f(\mathbf{a}_1,b_1) + (1-\lambda) f(\mathbf{a}_2,b_2).
\end{aligned}
$$
So, I think this can be applied to my problem, as the function $h(\mathbf{a},b,\mathbf{x})=\mathbf{a}\cdot\mathbf{x}+b$ is convex (actually it is affine with respect to $\mathbf{a}$, $b$) with respect to $\mathbf{a}$, $b$.

A Counterexample?
Let $h(\mathbf{a},b,\mathbf{x})=\mathbf{a}\cdot\mathbf{x}+b+1$, which is affine w.r.t. $\mathbf{a},b$ over the domain $\Omega=\{\mathbf{x}\in\mathbb{R}^n \mid \mathbf{a}\cdot\mathbf{x}+b+1 \geq 0\}$. Also, let $g$ be the probability density function of an $n$-dimensional Gaussian with mean $\mu\in\mathbb{R}^n$ and covariance matrix $\Sigma\in\mathbb{S}_{++}^n$ (symmetric, positive-definite, $n\times n$). Then, the function
$$
f(\mathbf{a},b)=\int_{\Omega}\! h(\mathbf{a},b,\mathbf{x})g(\mathbf{x})  \,\mathrm{d}\mathbf{x}
$$
is argued to be convex from the above "Theorem", but it is evaluated as follows
$$
f(\mathbf{a},b)
=
\frac{\mathbf{a}\cdot\mu+b+1}{2}
\left[
1 - \operatorname{erf}\left(-\frac{\mathbf{a}\cdot\mu+b+1}{\sqrt{2\mathbf{a}^T\Sigma\mathbf{a}}}\right)
\right]
+
\frac{\sqrt{\mathbf{a}^T\Sigma\mathbf{a}}}{2\pi}
\exp\left(
-\frac{(\mathbf{a}\cdot\mu+b+1)^2}{2\mathbf{a}^T\Sigma\mathbf{a}}\right),
$$
which obviously is not convex in $(\mathbf{a},b)$.
Where is the mistake?
