Problem 4

Exercise 9.4 (Slicing the domain of an integral)   Let $f_1,f_2,\dots$ and $g$ be functions in $L^1(\mathbb{R})$ and $E_n:=\{x\in\mathbb{R} \;\vert\; \vert f_n(x)\vert>\vert g(x)\vert\}$. Suppose $f_n\to g$ pointwise almost everywhere and

$$\lim_{n\to\infty} \int_{E_n}\vert f_n\vert = 0$$ (9.20)

Prove that

$$\lim_{n\to\infty} \int_{\mathbb{R}}\vert f_n-g\vert = 0$$ (9.21)

Proof. Break up the domain of the integral

$$\int_{\mathbb{R}}\vert f_n-g\vert = \int_{\mathbb{R}\setminus E_n}\vert f_n-g\vert + \int_{E_n}\vert f_n-g\vert$$ (9.22)

On the set $E_n$, we have $\vert g\vert<\vert f_n\vert$, so

$$\lim_{n\to\infty} \int_{E_n}\vert f_n-g\vert \leq \lim_{n\to\infty} \int_{E_n}2\vert f_n\vert = 0$$ (9.23)

On the complement, we have $\vert f_n\vert<\vert g\vert$, so that a dominating function exists, and DCT may be applied:

$$\lim_{n\to\infty}\int_{\mathbb{R}\setminus E_n}\vert f_n-g\vert =\lim_{n\to\infty}\int_{\mathbb{R}}\vert f_n-g\vert\cdot 1_{\mathbb{R}\setminus E_n}$$ (9.24)

$$=\int_{\mathbb{R}}\lim_{n\to\infty}\vert f_n-g\vert \cdot 1_{\mathbb{R}\setminus E_n}$$ (9.25)

$$\leq\int_{\mathbb{R}}\lim_{n\to\infty}\vert f_n-g\vert$$ (9.26)

$$= 0$$ (9.27)

Therefore

$$\lim_{n\to\infty} \int_{\mathbb{R}}\vert f_n-g\vert = 0$$ (9.28)

$\qedsymbol$