Problem 4

Exercise 8.4 (Convergence in measure metric)   Let $f_n : E\to\mathbb{R}$ be a sequence of measurable functions where $\mu(E)<\infty$. Then $f_n\to 0$ in measure if and only if

$$\lim_{n\to\infty} \int \frac{\vert f_n\vert}{1+\vert f_n\vert} = 0$$ (8.11)

Proof. Suppose the limit is zero. Let $\epsilon>0$ be given. Select $N$ such that

$$n\geq N \implies \int \frac{\vert f_n\vert}{1+\vert f_n\vert} < \epsilon^2$$ (8.12)

We are going to prove that the following set has small measure for such $n$.

$$F = \{ x\in E \;\vert\; \vert f_n(x)\vert>\epsilon\}$$ (8.13)

Measure by integrating. If $x\in F$, then $1<\vert f_n(x)\vert/\epsilon$, so we have

$$\mu(F) = \int_Fd\mu < \int_F \frac{\vert f_n(x)\vert}{\epsilon}$$ (8.14)

$$= \frac{1}{\epsilon}\int_F \vert f_n\vert$$ (8.15)

$$\leq \frac{1}{\epsilon}\int_F \frac{\vert f_n\vert}{1+\vert f_n\vert}$$ (8.16)

$$\leq \frac{1}{\epsilon}\int \frac{\vert f_n\vert}{1+\vert f_n\vert}$$ (8.17)

$$\leq \frac{1}{\epsilon} \epsilon^2$$ (8.18)

$$\leq \epsilon$$ (8.19)

Suppose convergence in measure holds. Let $\epsilon>0$ be given and define $\epsilon'=\epsilon/(2\mu(E))$ and define

$$F = \{x \in E \;\vert\; \vert f_n(x)\vert>\epsilon'\}$$ (8.20)

Break up the norm integral:

$$\int \frac{\vert f_n\vert}{1+\vert f_n\vert} = \int_{E\setminus F... ...rt f_n\vert}{1+\vert f_n\vert} + \int_F \frac{\vert f_n\vert}{1+\vert f_n\vert}$$ (8.21)

If $x \in E\setminus F$, then

$$\vert f_n(x)\vert \leq \epsilon' \implies \frac{\vert f_n(x)\vert}{1+\vert f_n(x)\vert}<\epsilon'$$ (8.22)

Also, $\vert f_n\vert/(1+\vert f_n\vert)<1$, so each integral can be bounded

$$\int \frac{\vert f_n\vert}{1+\vert f_n\vert} < \mu(E\setminus F)\epsilon' + \mu(F)$$ (8.23)

By convergence in measure, select $N$ such that $n\geq N$ implies $\mu(F)<\epsilon/2$. Then we are done. $\qedsymbol$