Convergence in $\mathbb{L}_1$ implies convergence "perturbed" conditional expectations

Question

Consider a sequence of conditional pdf's $p_n(y | x)$ on a Polish space $X \times Y$, endowed with its Borel sigma algebra. Suppose, as $n\rightarrow \infty$, in $\mathbb{L}_1$ (the following densities can be seen as mixtures of normals with weights functions $\omega_{k,n}(X)$, $\omega^*_k(X)$ that sum up to 1 almost surely for fixed $X$ and $n$):

$$ \sum_{k=1}^{\infty} \omega_{k,n} (X) \frac{1}{\sqrt{2 \pi \sigma^2_k}} \exp\left( -\frac{(y - \beta_{0,k,n} - \beta_{1,k,n} X)^2}{2 \sigma^2_k} \right) = p_n(y | X) \stackrel{\mathbb{L}_1}{\rightarrow} p_0(y | X) = \sum_{k=1}^{\infty} \omega^*_k (X) \frac{1}{\sqrt{2 \pi \sigma^2_{k,*}}} \exp\left( -\frac{(y - \beta^*_{0,k} - \beta^*_{1,k} X)^2}{2 \sigma^2_{k,*}} \right) $$

is then true that for a sequence of measurable function $f_n \stackrel{\mathbb{L}_1}{\rightarrow} f_0$ also the "perturbed" conditional expectations converge as follows:

$$ \sum_{k=1}^{\infty} \omega_{k,n} (X) (\beta_{0,k,n} + \beta_{1,k,n} (X+f_n(X))) \stackrel{\mathbb{L}_1}{\rightarrow} \sum_{k=1}^{\infty} \omega^*_k (X) (\beta^*_{0,k} + \beta^*_{1,k} (X+f_0(X))) $$

Answer 1

I don't want to come across as unnecessarily pedantic without being helpful, so I'm guessing what the question really means. Under my guessed assumptions, the answer is in general no even for $f_n = f_0 = 0$ for all $n \in \mathbb{N}$.

For $n \in \mathbb{N}$, let $\beta_{0, 1, n} = 0$, $\beta_{0, 2, n} = n$, $\beta_{1, k, n} = 0$ and $\sigma_k^2 = 1$ for all $k \in \mathbb{N}$. Also let $\omega_{1, n} = 1 - \frac{1}{n}$, $\omega_{2, n} = \frac{1}{n}$ and $\omega_{k, n} = 0$ for $k \geq 2$.

Regarding the limit, let $\beta_{0, 1}^* = 0$ and $\beta_{1, k}^* = 0$ and $\sigma_k^2 = 1$ for all $k \in \mathbb{N}$. Also let $\omega_1^* = 1$.

Then $$ \int \, \nu(\mathrm{d}x) \, \Vert p_n(\cdot \vert x) - p_0(\cdot \vert x) \Vert_{L^1} = \int \, \nu(\mathrm{d}x) \, \int_{\mathbb{R}} \, \mathrm{d}y \, \vert p_n(y \vert x) - p_0(y \vert x)\vert \leq \frac{2}{n} \xrightarrow[n \to \infty]{} 0 $$ but $$ \int \, \nu(\mathrm{d}x) \, \left\vert \omega_{1, n} \beta_{0, 1, n} + \omega_{2, n} \beta_{0, 2, n} - \omega_{1}^* \beta_{0, 1}^* \right\vert = \int \, \nu(\mathrm{d}x) \, \left\vert 0 + n \frac{1}{n} - 0\right\vert = 1, $$ hence the desired convergence does not hold.