Product of sequences convergence in $L^1$ and pointwise

Question

Let $a_n(x)=f_n(x)\cdot g(x)$ where $g(x)$ is lebesgue integrable $\mathscr{L}^1(\lambda)$. Meaning

$$g(x)\quad \text{measurable and} \int |g|d\lambda < \infty$$

Suppose $f_n$ converges pointwise to $f=0$ but not $L^1$.

I am confused because in this question Lebesgue integrable implies bounded a.e it seems this doesn't necessarily mean $g$ is bounded.

Does $\int |g|d\lambda < \infty $ imply $g$ is bounded? If $g$ is bounded then $a_n$ converges point wise to $0$ but not $L^1$ because $f_n$is not Cauchy.

Answer 1

Integrable does not imply function-boundedness. Consider $(0,1]$ with the usual Lebesgue measure, then $\displaystyle\int_{(0,1]}\dfrac{1}{x^{1/2}}dx<\infty$ but apparently $1/x^{1/2}$ is not bounded on $(0,1]$.

Answer 2

No, $\int |g| d\lambda < \infty$ does not imply that g is bounded. Notice that if we take $g\in \mathscr{L}^1(\lambda)((a,b))$, and an enumeration $(q_n)_n$ of the rational numbers in $(a,b)$, then by replacing the value of $g$ at $q_n$ with $n$ we obtain a new function $G$ which is equivalent to $g$ in $\mathscr{L}^1(\lambda)((a,b))$ (they agree almost everywhere in $(a,b)$) but $G$ is unbounded.