Let $\;f_n\;$ be a sequence of functions such that $\;f_n \in L^2((l_{-}^{f_n},l_{+}^{f_n}),\mathbb R^m)\;$ where $\;-\infty\le l_{-}^{f_n} \lt l_{+}^{f_n} \le +\infty\;$. If I knew $\;\lim_{n \to \infty} l_{\pm}^{f_n}=l_{\pm}^{\infty}\;$, would it be possible to claim that $\;f_n \in L^2((l_{-}^{\infty},l_{-}^{\infty}),\mathbb R^m)\;$?
My approach:
Since $\;f_n \in L^2((l_{-}^{f_n},l_{+}^{f_n}),\mathbb R^m)\;$, it holds $\; \int_{l_{-}^{f_n}}^{l_{+}^{f_n}} {f_n(x)}^2 \;dx \lt \infty\;$. If I let $\;n\to \infty\;$, I get confused because in this case (I believe) I get $\; \int_{l_{-}^{\infty}}^{l_{+}^{\infty}} \lim {f_n(x)}^2 \;dx\;$. But all I want is $\; \int_{l_{-}^{\infty}}^{l_{+}^{\infty}} {f_n(x)}^2 \;dx\;$.
What am I missing? Is my initial question wrong? Any help would be valuable!
Thanks in advance!!!
You can extend your functions $f_n$ with $0$ on $(l_-^\infty,l_+^\infty)\setminus (l_-^{f_n},l_+^{f_n})$.
Then
$$ \; \int_{l_-^\infty}^{l_+^\infty} {f_n(x)}^2 \;dx \leq \; \int_{\min(l_-^\infty,l_-^{f_n})}^{\max(l_+^\infty,l_+^{f_n})} {f_n(x)}^2 \;dx = \; \int_{l_{-}^{f_n}}^{l_{+}^{f_n}} {f_n(x)}^2 \;dx \lt \infty\;. $$