$L^{p}$ convergence equivalent condition I have to show that for $p\in[0,\infty)$, $f_{n},f\in L^{p}(\mathbb{R})$, (i) $f_n\rightarrow f$ in $L^{p}([-N,N])$ for all $N\in \mathbb{N}$ and (ii) $\lvert\lvert f_{n}\rvert\rvert \rightarrow \lvert\lvert f\rvert\rvert$ implies $f_{n}\rightarrow f$ in $L^{p}(\mathbb{R})$.
My approach was to first deconstruct $\lvert\lvert f_{n}-f\rvert\rvert_{p}^{p}$ as
$$
\lvert\lvert f_{n}-f\rvert\rvert_{p}^{p}=\int_{-\infty}^{-N}\lvert f_{n}-f\rvert^{p}+\int_{-N}^{N}\lvert f_{n}-f\rvert^{p}+\int_{N}^{\infty}\lvert f_{n}-f\rvert^{p}.
$$
(The integral is taken w.r.t. the Lebesgue measure). By (i) we know that the above decomposition is valid for all $N\in \mathbb{N}$ and as we take the limit $n\rightarrow \infty$, the middle term vanishes. The problem is how to get an upper bound $\epsilon_{n}$ for the left and right integral s.t. $\epsilon_{n}\rightarrow 0$ as $n\rightarrow \infty$. I don't see how we could use (ii) for that. I tried to show that  for all $n$ sufficiently large we get an $N^{*}$ such that the left and right integral are bounded and that this bound approaches $0$ as $n \rightarrow \infty$, but didn't succeed. Could anyone help?
 A: Let $\epsilon > 0$, and choose $N$ so large that $\|f\|_{L^p(\mathbb R\setminus [-N,N])} < \epsilon$. (Why does such an $N$ exist?)
Note that (i) $\|f-f_n\|_{L^p([-N,N])}\to 0$ implies $\|f_n\|_{L^p([-N,N])}\to \|f\|_{L^p([-N,N])}$. Then (ii) $\|f_n\|_p\to \|f\|_p$ and $\|f_n\|_{L^p([-N,N])}\to \|f\|_{L^p([-N,N])}$ together imply
$$
\|f_n\|_{L^p(\mathbb R\setminus[-N,N])}^p = \|f_n\|_p^p - \|f_n\|_{L^p([-N,N])}^p\to  \|f\|_{L^p(\mathbb R\setminus [-N,N])}^p. 
$$
So
$$
\|f-f_n\|_p^p = \|f-f_n\|_{L^p([-N,N])}^p + \|f-f_n\|_{L^p(\mathbb R\setminus [-N,N])}^p,
$$
and sending $n\to\infty$ the limit superior of the right-hand side is bounded by $(2\|f\|_{L^p(\mathbb R\setminus [-N,N])})^p < (2\epsilon)^p$.
A: Choose $M>0$ such that $\int|f_{n}|^{p}\leq M$ for all $n$. (This
is possible becuase $\int|f_{n}|^{p}\rightarrow\int|f|^{p}<\infty$).
Note that we also have $\int|f|^{p}\leq M$. Let $\varepsilon>0$
be arbitrary. Since $f\in L^{p}$, we may choose $N\in\mathbb{N}$
such that $\int_{[-N,N]^{c}}|f|^{p}<\varepsilon$ (This follows from
Dominated Convergence Theorem by observing that $1_{[-N,N]^{c}}(x)|f(x)|^{p}\rightarrow0$
as $N\rightarrow\infty$ and that it is dominated by the integrable
function $|f|^{p}$). Since $\int_{[-N,N]}|f_{n}-f|^{p}\rightarrow0$
as $n\rightarrow\infty$, we may choose $N_{1}\in\mathbb{N}$ such
that $\int_{[-N,N]}|f_{n}-f|^{p}<\varepsilon^{p}$ whenever $n\geq N_{1}$.
Choose $N_{2}\in\mathbb{N}$ such that $\left|\int|f_{n}|^{p}-\int|f|^{p}\right|<\varepsilon$
whenever $n\geq N_{2}$.
Let $n\geq\max(N_{1},N_{2})$. We go to prove that $\int_{[-N,N]^{c}}|f_{n}|^{p}$
is small. We have estimation:
\begin{eqnarray*}
 &  & \int_{[-N,N]^{c}}|f_{n}|^{p}+\int_{[-N,N]}|f_{n}|^{p}\\
 & = & \int|f_{n}|^{p}\\
 & \leq & \int|f|^{p}+\varepsilon\\
 & = & \int_{[-N,N]^{c}}|f|^{p}+\int_{[-N,N]}|f|^{p}+\varepsilon.\\
 & \leq & 2\varepsilon+\int_{[-N,N]}|f|^{p}.
\end{eqnarray*}
Therefore, $\int_{[-N,N]^{c}}|f_{n}|^{p}\leq2\varepsilon+\int_{[-N,N]}|f|^{p}-\int_{[-N,N]}|f_{n}|^{p}$.
To simplify notation, we denote $||g||=\left\{ \int_{[-N,N]}|g|^{p}\right\} ^{\frac{1}{p}}$
for any measurable function $g:\mathbb{R}\rightarrow\mathbb{R}$.
We have estimation:
\begin{eqnarray*}
 &  & ||f||\\
 & = & ||(f-f_{n})+f_{n}||\\
 & \leq & ||f-f_{n}||+||f_{n}||\\
 & \leq & \varepsilon+||f_{n}||.
\end{eqnarray*}
Let $\theta:[0,\infty)\rightarrow\mathbb{R}$ be defined by $\theta(x)=x^{p}$.
By mean-value theorem, we have that
\begin{eqnarray*}
\theta(||f||)-\theta(||f_{n}||) & = & \theta'(\xi)\left\{ ||f||-||f_{n}||\right\} \\
 & \leq & \varepsilon\theta'(\xi)
\end{eqnarray*}
where $\xi$ is a number between $||f||$ and $||f_{n}||$, and hence
$\xi\leq M^{\frac{1}{p}}$. It follows that $\theta'(\xi)=p\xi^{p-1}\leq p\cdot M^{\frac{p-1}{p}}.$
Therefore,
\begin{eqnarray*}
\int_{[-N,N]^{c}}|f_{n}|^{p} & \leq & 2\varepsilon+||f||^{p}-||f_{n}||^{p}\\
 & \leq & 2\varepsilon+\varepsilon pM^{\frac{p-1}{p}}.
\end{eqnarray*}
Furthermore, we have that
\begin{eqnarray*}
\left\{ \int_{[-N,N]^{c}}|f_{n}-f|^{p}\right\} ^{\frac{1}{p}} & \leq & \left\{ \int_{[-N,N]^{c}}|f_{n}|^{p}\right\} ^{\frac{1}{p}}+\left\{ \int_{[-N,N]^{c}}|f|^{p}\right\} ^{\frac{1}{p}}\\
 & \leq & \left\{ 2\varepsilon+\varepsilon pM^{\frac{p-1}{p}}\right\} ^{\frac{1}{p}}+\varepsilon^{\frac{1}{p}}.
\end{eqnarray*}
Finally, for $n\geq\max(N_{1},N_{2})$, we have estimation
\begin{eqnarray*}
 &  & \int|f_{n}-f|^{p}\\
 & = & \int_{[-N,N]}|f_{n}-f|^{p}+\int_{[-N,N]^{c}}|f_{n}-f|^{p}\\
 & \leq & \varepsilon^{p}+\left[\left\{ 2\varepsilon+\varepsilon pM^{\frac{p-1}{p}}\right\} ^{\frac{1}{p}}+\varepsilon^{\frac{1}{p}}\right]^{p}.
\end{eqnarray*}
It follows that $\int|f_{n}-f|^{p}\rightarrow0$ as $n\rightarrow\infty$.
A: We can show it using the following facts.

*

*If $\left(g_n\right)_{n\geqslant 1}$ is a sequence such that $g_n\to g$ almost everywhere and $\int \lvert g_n\rvert^p\to\int \lvert g\rvert^p$, then $\int \lvert g_n-g\rvert^p\to 0$.


*If $(f_n)_{n\geqslant 1}$ satisfies the conditions of the opening post, then there exists a subsequence $(g_n)=(f_{\varphi(n))})$ where $\varphi\colon\mathbb N\to\mathbb N$ is increasing, such that $\int \lvert g_n-f\rvert^p\to 0$.
In order to show fact one, we can use Fatou's lemma to the sequence $(h_n)$ given by
$$
h_n=
\begin{cases}
\lvert f\rvert^p+\lvert f_n\rvert^p-\lvert f_n-f\rvert^p&\mbox{ if }0<p\leqslant 1,\\
2^{p-1}\lvert f\rvert^p+2^{p-1}\lvert f_n\rvert^p-\lvert f_n-f\rvert^p&\mbox{ if } p\geqslant 1.
\end{cases}
$$
For fact 2., we use the fact that convergence in $L^p$ implies the almost sure convergence of a subsequence combined with a diagonal extraction argument.
To conclude, fact 2. applied to a subsequence of $(f_n)$, say $(f_{\psi(n))})$  show that there exists a further subsequence $(f_{\varphi\circ \psi(n)})$ such that $\int\left\lvert f_{\varphi\circ \psi(n)}-f\right\rvert^p\to 0$.
