TITLE: Showing integrability of a locally integrable function on a bounded domain under some additional assumptions
QUESTION [5 upvotes]: Suppose $\Omega\subset \mathbb{R}^3$ is a smooth and bounded domain, and $f:\Omega\to[0,\infty]$ is a given function which is finite almost everywhere and satisfies

Assumption A: For all $g\in C_0^1(\Omega)$ we have the product $fg\in L^1(\Omega)$.  (Here $C_0^1(\Omega)$ refers to functions which are continuously differentiable in $\Omega$ and extend continuously to $0$ on $\partial\Omega$).

Question 1:  Can we show that $f\in L^1(\Omega)$?
Question 2:  Does the answer to Question 1 change if we include some or all of the following assumptions:

Assumption B:  $f$ possesses a weak derivative which is finite almost everywhere in $\Omega$;

Assumption C:  There exists a nonnegative function $f_0 \in H^2(\Omega)\cap C(\bar{\Omega})$ such that $f-f_0=0$ (in the sense of trace) on $\partial\Omega$;

Assumption D: There exists a nonnegative function $h\in H^1(\Omega)$ such that $h$ is nonzero almost everywhere in $\Omega$ and $f=-\ln h$ in $\Omega$.


Note:  Assumption D more or less implies Assumption B.  I wrote them separately in the hopes of formulating the problem as simply as possible.
Notation: Here $H^k$ is the standard Sobolev space notation for $W^{k,2}$.
9/14/20 Edit:
Question 1 has been answered in the affirmative.  I additionally pose the following
Question 3:  Answer  Questions 1 and 2 in the case that Assumption A is replaced by

Assumption A': $f\in L^1_{\text{loc}}(\Omega)$.

REPLY [7 votes]: Let $(g_k)_{k\ge0}$ be a sequence of smooth functions  such that    $g_k(x)=1$ if  $\text{dist}(x,\partial\Omega)\ge 2^{-k}$,
$g_k (x)=0$ if  $\text{dist}(x,\partial\Omega)\le 2^{-k-1}$ and $0\le g_k\le 1$ everywhere.
To prove the affirmative answer to Question 1 by contrapositive, let $f\not\in L^1(\Omega)$ be given: we want to find $g\in C^1_0(\Omega)$ such that $fg\not\in L^1(\Omega)$. We can assume $fg_k\in L^1(\Omega)$ for all $k$, otherwise we are  done with $g=g_k$ for some $k$. Then $ \int_\Omega fg_k$ is an increasing sequence of positive real numbers, that diverges to $+\infty$, for if it were bounded, $f\in L^1(\Omega)$ by Beppo Levi's theorem. So  for some subsequence $(g_{k_j})_j$ we have $ \int_\Omega fg_{k_{j+1}}\ge \int_\Omega fg_{k_j}+1  $, that is $ \int_\Omega f(g_{k_{j+1}}-g_{k_j})\ge 1$. For all $j$ the function $g_{k_{j+1}}-g_{k_j}$ is bounded between $0$ and   $1$, and supported in the set  $\big\{ 2^{-k_{j+1}-1}\le \text {dist}(x,\partial\Omega)\le 2^{-k_j}\big\}$.
But then $g:=\sum _{j\ge1} \frac{g_{k_{j+1}}-g_{k_j} }j$ is a locally finite sum of smooth functions, hence smooth in $\Omega$; clearly $g(x)\to0$ for $x\to\partial\Omega$, and, again by Beppo Levi's theorem, $\int_\Omega fg\ge \sum_{j\ge1}\frac1j=+\infty$.