$L^p$ convergence implies $\int_{\mathbb{R}^d \setminus E} |f_n|^p < \epsilon$ for all $n$.

Question

Question. Suppose $f_n, f \in L^p(\mathbb{R}^d)$ such that $f_n \to f$ in $L^p$-norm. Is it necessarily true that for all $n \in \mathbb{N}$, and any $\epsilon > 0$, there exists a $E \subset \mathbb{R}^d$ with finite measure $\mu(E) < \infty$ such that $\int_{\mathbb{R}^d \setminus E} |f_n|^p < \epsilon$.

I believe this is true. However, I'm a little unsure of how to rigorously prove.

What I have: Since $E$ is measurable, then $F = \mathbb{R}^d \setminus E$ is also measurable. Then

$$\int_F |f_n|^p = \int_F |f_n - f + f|^p \leq \int_F |f_n - f|^p + \int_F |f|^p.$$

Since $f_n \to f$ in $L^p$-norm, for each $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that $$\int_F |f_n - f|^p < \frac{\epsilon}{2},$$

whenever $n \geq N$. Further, $f$ is integrable. Meaning there exists some $\epsilon > 0$ such that $$\int_F |f|^p < \frac{\epsilon}{2}.$$

Thus, for $n \geq N$,

$$\int_F |f_n|^p < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$

But how does one proceed when $n < N$?

Answer 1

Let us argue by contradiction. Suppose it does not hold. That means there is an $\varepsilon > 0$ such that for all finite measurable $E \subset \mathbb{R}^d$ there exists some $n\in\mathbb{N}$ so that $\int_{E^c} |f_n|^p > \varepsilon$.

Choose $\varepsilon > 0$. Since $f\in L^p$ we know there is some finite and measurable $F \subset \mathbb{R}^d$ such that

$$ \int_{F^c} |f|^p < \frac{\varepsilon}{2}. $$

Now we proceed as you do. Since $f_n\rightarrow f$ in $L^p$, there is some natural number $N$ such that for every $n\geq N$

$$ \int_{F^c} |f_n|^p \leq \int_{F^c} |f_n-f|^p + \int_{F^c} |f|^p < \frac{\varepsilon}{2}+\frac{\varepsilon}{2} = \varepsilon. $$

Therefore we already know that our new claim is not true for any $n\geq N$. But the remaining $N-1$ functions are a finite amount, so we can simply define finite and measurable sets $E_n \subset \mathbb{R}^d$ such that

$$ \int_{E_n^c} |f_n|^p < \varepsilon $$

for all $1\leq n< N$. Since we can choose whatever $\varepsilon > 0$ and repeat the same argument the new claim is false and the answer to your question is a positive answer.