Problem 4

Exercise 1.4 (DCT for convergence in measure)   Let $f_n$ be a sequence of measurable functions converging in measure to $f$ and pointwise bounded by $\vert f_n(x)\vert\leq g(x)$ where $g\in L^1$. Then

$$f \in L^1 \quad \lim_{n\to\infty} \int_X \vert f_n-f\vert d\mu = 0$$ (1.20)

Proof. To proceed, we will prove these in the case that $\mu(X)<\infty$ and then extend the result.

To show $f$ is simply integrable, we show $\vert f(x)\vert \leq g(x)$ almost everywhere. For each $\epsilon>0$, define

$$E_\epsilon^n := \{ x\in X \;\vert\; \vert f(x)-f_n(x)\vert \geq \epsilon \}$$ (1.21)

Define

$$S = \bigcap_{n=1}^\infty E_\epsilon^n$$ (1.22)

Then

$$\mu(S) \leq \mu(E_\epsilon^n) \quad \forall n$$ (1.23)

The definition of convergence in measure guarantees $\mu(S)=0$. Therefore

$$\int_X\vert f(x)\vert = \int_{X\setminus S} \vert f(x)\vert$$ (1.24)

The integrand is then bounded by comparing

$$\vert f(x)\vert \leq \vert f(x)-f_n(x)\vert + \vert f_n(x)\vert$$ (1.25)

If $x\in X\setminus S$, then $x\notin E_\epsilon^n$, so $f$ is bounded by $\epsilon$ and the dominating function

$$\vert f_n(x)-f(x)\vert<\epsilon \implies \vert f(x)\vert \leq \epsilon + g(x)$$ (1.26)

Therefore,

$$\int\vert f(x)\vert \leq \int \epsilon+g(x) = \epsilon\mu(X) + \int g(x)$$ (1.27)

Letting $\epsilon\to 0$, we find the desired result.

Now that $f\in L^1$, let us show

$$\lim_{n\to\infty} \int_X \vert f_n(x)-f(x)\vert=0$$ (1.28)

Let $\epsilon>0$ be given and define $E_\epsilon^n$ as above. Select $\delta>0$ so that $\mu(B)<\delta$ implies

$$\int_B 2g < \epsilon$$ (1.29)

We are ready for the limit

$$\lim_{n\to\infty} \int_X \vert f_n(x)-f(x)\vert = \lim_{n\to\infty} \left( \int_{X\setminus E_\epsilon^n} \vert f_n(x)-f(x)\vert + \int_{E_\epsilon^n} \vert f_n(x)-f(x)\vert \right)$$ (1.30)

$$\leq \lim_{n\to\infty} \left( \int_{X\setminus E_\epsilon^n} \vert f_n(x)-f(x)\vert + \int_{E_\epsilon^n} 2g \right)$$ (1.31)

Select $N$ such that $n\geq N$ implies $\mu(E_\epsilon^n)<\delta$. This bounds the second integral by $\epsilon$ from the integral estimate. The first integral has an easy bound by convergence in measure:

$$\int_{X\setminus E_\epsilon^n} \vert f_n(x)-f(x)\vert \leq \int_{... ...silon^n} \epsilon \leq \epsilon\mu(X\setminus E_\epsilon^n) \leq \epsilon\mu(X)$$ (1.32)

Since $\epsilon>0$ was arbitrary, we have

$$\lim_{n\to\infty} \int_X \vert f_n(x)-f(x)\vert=0$$ (1.33)

Now suppose $\mu(X)=\infty$. Let $\epsilon>0$ be given. Select $F\subseteq X$ such that $\mu(F)<\infty$ and

$$\int_{X\setminus F} 2g < \epsilon$$ (1.34)

by the integrability of $2g$. Then

$$\int_X \vert f_n(x) - f(x)\vert = \int_{X\setminus F} \vert f_n(x) - f(x)\vert + \int_F\vert f_n(x) - f(x)\vert$$ (1.35)

$$\leq \int_{X\setminus F}2g + \int_F\vert f_n(x)-f(x)\vert$$ (1.36)

$$\leq \epsilon + \int_F\vert f_n(x)-f(x)\vert$$ (1.37)

Since $F$ has finite measure, we can apply the previous result to show

$$\lim_{n\to\infty} \int_X \vert f_n(x) - f(x)\vert \leq \epsilon + \lim_{n\to\infty}\int_F\vert f_n(x)-f(x)\vert \leq \epsilon$$ (1.38)

The selection of $\epsilon$ was arbitrary, so we know that this limit equals zero. $\qedsymbol$

For an extra goodie, we look at how the convergence in measure metric $\rho$ might have been used to solve this problem. Let $\epsilon>0$ be given. Bound the integral for a fixed $n$

$$\int_X \vert f_n(x) - f(x)\vert = \int_X \frac{\vert f_n(x)-f(x)\vert}{1+\vert f_n(x)-f(x)\vert} \left[1+\vert f_n(x)-f(x)\vert\right]$$ (1.39)

$$= \operatorname{ess}\sup \frac{\vert f_n-f\vert}{1+\vert f_n-f\vert} \times \int_X \left[1+\vert f_n(x)-f(x)\vert\right]$$ (1.40)

$$\leq \operatorname{ess}\sup \frac{\vert f_n-f\vert}{1+\vert f_n-f\vert} \times (\mu(X)+\Vert 2g\Vert)$$ (1.41)

This essential supremum exists because the ratio in the integrand is bounded by one, so we can find the $\operatorname{ess}\sup$ as a limit of $L^p$ norms. Select $p$ such that

$$\operatorname{ess}\sup \frac{\vert f_n-f\vert}{1+\vert f_n-f\vert} = \left(\int \left(\frac{\vert f_n-f\vert}{1+\vert f_n-f\vert}\right)^p \right)^{1/p} + O(\epsilon/(\mu(X)+\Vert 2g\Vert))$$ (1.42)

$$\leq \left(\rho(f_n,f)\right)^{1/p} + O(\epsilon/(\mu(X)+\Vert 2g\Vert))$$ (1.43)

Combining these estimates shows

$$\int_X \vert f_n(x) - f(x)\vert \leq \rho(f_n,f)^{1/p}(\mu(X)+\Vert 2g\Vert) + O(\epsilon)$$ (1.44)

Select $N$ such that $n\geq N$ implies

$$\rho(f_n,f)= O((\epsilon/(\mu(X)+\Vert 2g\Vert))^p) \implies \rho(f_n,f)^{1/p}= O(\epsilon/(\mu(X)+\Vert 2g\Vert))$$ (1.45)

This result assumes $\mu(X)<\infty$, so we still have to extend the result as in the previous proof.