Chebyshev's Inequality in proof of Proposition 23 of Royden?

Question

This is from the 4th edition of Royden, on page 92.

Proposition 23. Let $f$ be a measurable function on $E$. If $f$ is integrable over $E$, then for each $\epsilon >0$, there is a $\delta >0$ for which $$\text{if } A\subset E \text{ is measurable and } m(A) < \delta, \text{ then } \int_A |f| < \epsilon. \tag{*}$$

Royden's proof:

We assume that $f \geq 0$ on $E$ and that $f$ is integrable over $E$. Then by additivity over domains of integration and Chebyshev's Inequality, if $A\subset E$ is measurable and $c > 0$, then $$ \int_A f = \int_{\{x\in A\, \mid\, f(x) < c\} } f + \int_{\{x\in A\, \mid\, f(x) \geq c\} } f \leq c \cdot m(A) + \frac{1}{c} \int_E f. \tag{1}$$

Choose $c >0$ such that $1/c \cdot \int_E f < \epsilon /2$. Then $$\int_A f < c \cdot m(A) + \epsilon /2. $$ Define $\delta = \epsilon / 2c$. Then (*) holds for this choice of $\delta$.

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I'm having trouble working out this last inequality. So I have $$\int_{\{x\in A\, \mid\, f(x) < c\} } f = \int_A f \cdot \chi_{\{x\in A\, \mid\, f(x) < c\} } < \int_A c = c \cdot m(A),$$ which takes care of the first part, but I can't see how to use Chebyshev's Inequality to get $$\int_{\{x\in A\, \mid\, f(x) \geq c\} } f \leq \frac{1}{c} \int_E f,$$ since Chebyshev's inequality would give us $$\int_A \chi_{\{x\in A: f(x) \geq c\}} = m\{x\in A: f(x) \geq c\} \leq \frac{1}{c} \int_A f $$ but we have a non-constant integrand $f$? How do we use Chebyshev's Inequality to get (1)?

Answer 1

You don't need Chebyshev's Inequality here. The proof is as follow:

Let $B_n=\{x:x\in E,\:n-1<f(x)\leqslant n\}$ and $C=\{x:x\in E,\:f(x)>n\}$. We have $$ \int_Efdm=\int_{\bigcup_{n=1}^{\infty}B_n}fdm=\sum_{n=1}^{\infty}\int_{B_n}fdm $$ Since $\sum_{n=1}^{\infty}\int_{B_n}fdm$ is absolute convergent, given $\epsilon>0$ there is a $N$ such that for any $n>N$ $$ \sum_{n=N+1}^{\infty}\int_{B_n}fdm=\int_{\bigcup_{n=N+1}^{\infty}B_n}fdm<\epsilon/2\tag{1} $$ Let $$ B=\bigcup_{n=1}^{N}B_n\quad\text{and}\quad C=\bigcup_{n=N+1}^{\infty}B_n=E-B $$ Take $\delta=\dfrac{\epsilon}{2(N+1)}$. Then for any $A\subset E$ such that $m(A)<\delta$, by $(1)$ $$ \int_Afdm=\int_{A\cap (B\cup C)}fdm=\int_{A\cap B}fdm+\int_{A\cap C}fdm\leqslant N\int_Adm+\int_Cfdm<N\dfrac{\epsilon}{2(N+1)}+\dfrac{\epsilon}{2}<\epsilon $$

Answer 2

Another way is via the Dominated Convergnce Theorem: $\lim_{c\to \infty} \int_{\{x\in A: f(x)\ge c\}} f(x)\,dx=\lim_{c\to \infty}\int_E 1_{\{f\ge c\}\cap A}f\,dx=0$ because the integrand $1_{\{f\ge c\}\cap A}f$ converges to $0$ a.e. and is dominated on $E$ by the integrable function $f$.