Problem:
Give an example of a sequence of non-negative functions $f_n$ tending to $0$ pointwise such that $\int f_n \to 0$, but there is no integrable function $g$ such that $f_n \leq g$ for all $n$.
My Ideas:
Take $([0,1], m, \mathcal B_{[0,1]})$ as the measure space. Let $f_n = \frac{1}{x}\cdot\chi_{\left(\frac{1}{n+1},\frac{1}{n}\right]}$ be our sequence of non-negative measurable functions. This sequence of functions does fit the conditions given. My logic is that since any function that dominates our functions $f_n$ would ultimately have to dominate the function $\frac{1}{x}$, which is not integrable, there can be no dominating function that is also integrable.
Does this look to be correct, and if so, is there any advice on how to make this argument tighter? Obviously missing the details showing that this sequence fits the conditions, but I have those already.
Yes, this is a fine argument; well done!
For something a bit simpler, perhaps study piecewise constant functions, such as
$$f_n(x) = n \chi_{(1/(n + 1), 1/n]}$$
Then we would have
$$\int f_n = n \left(\frac 1 n - \frac 1 {n + 1}\right) = n \cdot \frac{1}{n(n + 1)} = \frac 1 {n + 1}$$
On the other hand, a dominating function $g$ would necessarily satisfy $g \ge \sum_{n \ge 1} f_n$, but then $g$ is non-integrable since the harmonic series is divergent. (Alternatively, an application of the Dominated Convergence Theorem to this sequence easily implies the divergence of the harmonic series.)