If
$$f_n:X\rightarrow [0,\infty]$$
is a sequence of measurable functions and we know that $$\lim_{n\rightarrow \infty }\int_X f_n \,d\mu=0,\qquad \qquad \tag{$\star$}$$ then can we conclude that for any measurable set $Y\subset X$ we have $$\lim_{n\rightarrow \infty }\int_Y f_n \,d\mu=0$$ because $0\leq\int_Y f_n \,d\mu\leq\int_X f_n \,d\mu$?
Also, does the limit $(\star)$ imply that $\lim_{n\rightarrow \infty}f_n(x)=0$ almost everywhere?
On
The argument you give is correct: if a sequence of non-negative numbers converges to $0$ and another sequence is squeezed between it and $0$, then that other sequence also approaches $0$. And that other sequence is indeed squeezed between $0$ and the sequence that by hypothesis goes to $0$.
Here is an example of a sequence of non-negative-valued functions $f_n$ on $(0.1,\ 1)$ for which
$$ \lim_{n\to\infty}\int_{(0.1,\ 1)} f_n\,d\mu = 0\text{ but } \lim_{n\to\infty} f_n(x)\text{ fails to exist at every }x\in(0.1,\ 1). $$ $$ f_{2413}(x) = \begin{cases} 1 & \text{if }\frac{2413}{10000} \le x < \frac{2413+1}{10000}, \\[6pt] 0 & \text{otherwise}, \end{cases} $$ and similarly for all other indices $n$ (e.g. if the index is $46$ then it is the indicator function of the interval $[46/100,\ 47/100)$, etc.).
Yes (monotonicity helps). No (but maybe a subsequence does?). Respectively.