I am trying to understand how multiplying by or adding a Lebesgue integrable function to a sequence of functions affects the convergence
Say $g(x) \in \mathscr{L^1}(\lambda)$. That is $g(x)$ is measurable and $ \int g(x)d\lambda < \infty$. We have to sequences: $$a_n(x)=f_n(x)g(x)=n^{-1}\chi_{[0,n)}(x) \cdot g(x)$$
Here although $f_n(x)$ does not converge in $L^1$ it converges in $L^\infty$ which implies $a_n$ converges in $L^1$. $f_n$ converges point wise to $0$ but can we say anything about the point wise convergence of $a_n$. I know since $L^1$ is complete, $a_n$ is Cauchy and therefore has a convergent subsequence.
What about a sequence like $$b_n(x)=n^{-1}\chi_{[0,1/n)} +g(x)$$
What restrictions do we need for convergence in $L^p$ to mean point wise convergence?
For some sort of "multiplier", one goes with the Holder's inequality to see what further properties we can get: $\|uv\|_{L^{r}}\leq\|u\|_{L^{p}}\|v\|_{L^{q}}$, where $\dfrac{1}{r}=\dfrac{1}{p}+\dfrac{1}{q}$, $1\leq r,p,q\leq\infty$ with the convention that $\dfrac{1}{\infty}=0$, $\dfrac{1}{0}=\infty$.
For the $L^{p}$ convergence, this only implies that there is a subsequence who converges a.e., there is hardly anything to say about the a.e. convergence of the original sequence.