Let $f_n$ is a sequence of measurable functions on a measure space $(X,\mathcal{B},m)$ converging pointwise to a function $f$. Suppose that $f_n$ is integrable for all $n$ and $$\sup_{n}\int_{X}f_ndm<\infty,$$ is it true that $f$ is integrable?
I could conclude in the positive for monotonic sequences $f_n$, however I am unable to proceed further in this case.
On
As already said, if $f_n\ge0$ this is just Fatou's Lemma.
And in general the answer is no. But only because you asked the wrong question! Of course the answer is no if you're not assuming $f_n\ge0$; saying that the integrals are bounded says nothing about the size of $f_n$, the integral could be small because of cancellation.
For that matter if the integrals tend to $-\infty$ they satisfy your hypothesis; simply saying $f_n=-\chi_{[0,n]}$ gives a simple counterexample.
You "should" instead be assuming $$\sup_n\int|f_n|<\infty.$$Now Fatou's Lemma shows $\int|f|<\infty$, exactly as before.
Using Fatou's lemma, if the functions $f_n$ are positive you don't need to suppose they are monotonic: $\int_X fdm \leq \liminf_n \int_Xf_n \leq \sup_n \int_X f_n dm < \infty$
If you don't suppose the functions positive, then the limit function may not be integrable. Take, for instance, the measure space $(X,\mathcal{B},m)=([-1,1], \mathcal{B}_{[-1,1]}, \lambda|_{[-1,1]})$ where $\lambda$ is the Lebesgue measure and consider the sequence of functions $(f_n)_{n \geq 1}$ given by $f_n = \text{sign}(x)\frac{1}{|x|^{1-1/n}} = \text{sign}(x)|x|^{1/n-1}$. They are all Lebesgue integrable with $\int_{X}f_n dm = 0$, hence $\sup_n \int_X f_n dm = 0$, but the limit function is not Lebesgue integrable.