Does the following integral exist Let $(F_n : n ∈ \mathbb{N})$ be an increasing sequence in $\mathcal{B}(\mathbb{R})$ and $f$ be a measurable function defined on $\cup_{n∈\mathbb{N}}F_n$. Suppose $f$ is integrable on $F_n$ for every $n∈\mathbb{N}$ and $\lim_{n \to \infty} \int_{F_n}{f}dλ$ exists in $\mathbb{R}$. Does $\int_{\cup_{n∈\mathbb{N}}F_n}f dλ$ exist?
So far, I have used a definition I found online: Let $\delta>0$, $F_n=F$, and $\forall n>0$: $\int_F{|f_n|<\epsilon}$, whenever $\mu(F)<\delta$. Then using this definition: $\int_{\cup_{n=1}^{M}F}|f_n|\leq \sum_{n=1}^{M}{\int_F{|f_n|}} \leq \sum_{n=1}^{M} \epsilon = \epsilon M$. So, I guess the integral exists but I'm not sure. I might be completely wrong but that's my only solution so far. Any help is appreciated!

It's not true in general.
Let $F_{n}=[-n,n]$ and let $f(x)=x$ . Then $\int_{F_{n}} f\,d\lambda= 0$ for all $n$  and hence the limit $\lim_{n\to\infty}\int_{F_{n}} f\,d\lambda =0$ .
Now $\bigcup_{n\in\mathbb{N}}F_{n}=\mathbb{R}$ but $x$ is not integrable over the real line.
However it's true if $f$ is non-negative. In that case let $g_{n}=f\cdot\mathbf{1}_{F_{n}}$ .
Then $g_{n}$ is an increasing sequence of non-negative measurable functions and $\lim_{n\to\infty} g_{n}= f\cdot\mathbf{1}_{\{\bigcup_{n\in\mathbb{N}} F_{n}\}}$ .
So By Monotone Convergence Theorem you get your required result as $\displaystyle\lim_{n\to\infty}\int_{F_{n}}f\,d\lambda=\lim_{n\to\infty}\int_{\Bbb{R}} g_{n}\,d\lambda = \int_{\Bbb{R}} f\cdot\mathbf{1}_{\bigcup_{n\in\mathbb{N}}F_{n}}\,d\lambda = \int_{\bigcup_{n\in\mathbb{N}}F_{n}}f\,d\lambda$
