Now the same proof as in the case printed in the notes remains valid. Below is the solution of the modified version of example 1.
Idea. Since for each $x\in E$ the inequality $f(x)>0$ holds, we can make use of this inequality to describe the set $E$.
Assume $m(E)>0$. Now for each $x\in E$, $f(x)>0$, so there is an $n\in \N$ such that $f(x)>1/n$, i.e., there is an $n\in \N$ so that we have \[
x\in \left\{t\in E: f(t)>\frac{1}{n}\right\}=:E_n,
\] and thus $x\in \cupp_{n=1}^\infty E_n$. We conclude $E\subseteq \cupp_{n=1}^\infty E_n$. As by definition $E_n\subseteq E$ for each $n$, we have \[
E=\cupp_{n=1}^\infty E_n.\] By subadditivity of Lebesgue measure, \[0<m(E)\leq \sum_{n=1}^\infty m(E_n),\] so there must be an $n$ such that $m(E_n)>0$. Now we have \[
f > \frac{1}{n} \text{ on $E_n$} \quad \text{and}\quad f>0 \text{ on $E\setminus E_n$},
\] Therefore we have \[
\color{blue}f\chi_{E_n}+\color{green}f\chi_{E\setminus E_n} > \color{blue}{\frac{1}{n}}\chi_{E_n}+\color{green} 0\chi_{E\setminus E_n}\text{ on $E$}\quad \iff \quad f>\chi_{E_n}\text{ on $E$}.
\]
Integration on both sides yields\[\int_Ef(x)\,dm \ge \frac{1}{n}m(E_n)>0.\]
Important Consequence. Note that we have an important corollary here
Corollary. Suppose that $f:A\to[0,\infty)$ is measurable. If $\int_A f\,dm =0$, then $f=0$ a.e. on $A$.Proof. Define $E:=\{x\in A:f(x)>0\}$, then we have $f>0$ on $E$ (by definition) and since $E\subseteq A$ we have \[
0\leq \int_E f\, dm \leq \int_A f\, dm =0\implies \int_Ef\,dm =0.
\] Therefore by the contrapositive of the statement in Example 1, we have \[\int_E f\,dm =0\implies m(E)=0,\] we can conclude $f\leq 0$ a.e. on $A$. Since $f\ge 0$ on $A$, we have $f=0$ a.e. on $A$.$\qed$
Example 2.
Method 1. By AM-GM inequality $\frac{a+b+c+d}{4}\ge \sqrt[4]{abcd}$ for $a,b,c,d\ge 0$, we have \[
f\leq \frac{f^3+\sqrt[3]{f}+\sqrt[3]{f}+\sqrt[3]{f}}{4}=\frac{1}{4}f^3+\frac{3}{4}\sqrt[3]{f},
\] therefore \[
\int_\R f\, dm \leq \frac{1}{4}\int_\R f^3\,dm +\frac{3}{4}\int_\R\sqrt[3]{f}\,dm <\infty.
\]
Method 2. We want to compare $f$ with $\sqrt[3]{f}$ and $f^3$, but
$f\leq \sqrt[3]{f}$ only when $f(x)\in [0,1]\iff x\in f^{-1}[0,1]$; and
$f\leq f^3$ only when $f(x)\in [1,\infty)\iff x\in f^{-1}[1,\infty)$,
so \begin{align*}
\int_\R f\,dm &=\int_{f^{-1}[0,1]}f\,dm +\int_{f^{-1}(1,\infty)}f\,dm \\
&\leq \int_{f^{-1}[0,1]}\sqrt[3]{f}\,dm + \int_{f^{-1}(1,\infty)}f^3\,dm\\
&\leq \int_\R \sqrt[3]{f}\,dm +\int_\R f^3\,dm \\
&<\infty.
\end{align*}
Example 3. When $\alpha=1$, we need to prove \[
\limn \int_E \ln \brac{1+\frac{f(x)}{n}}^n\,dm(x) =\int_E f\,dm.
\] Note that for every $y\ge 0$, \[
\brac{1+\frac{y}{n}}^n \upto e^y,
\] we prove this in the remark below. Therefore we have \[
\ln \brac{1+\frac{f(x)}{n}}^n \upto f(x)
\] pointwise on $E$. Since $\ln \brac{1+\frac{f(x)}{n}}^n\ge 0$ on $E$ for every $n$, so we are done by using MCT.
Remark.
Method 1. This method is elementary. We fix a $y\in [0,\infty)$, by AM-GM inequality,
\[\bigg(\brac{1+\frac{y}{n}}^n\cdot 1\bigg)^{\frac{1}{n+1}}\leq \frac{n(1+\frac{y}{n})+1}{n+1}\iff \brac{1+\frac{y}{n}}^n \leq \brac{1+\frac{y}{n+1}}^{n+1}.\]
Method 2. This method is nonelementary, but very routine. Let's define $f(x)=(1+\frac{y}{x})^x$ and show that $f$ is increasing on $(0,\infty)$. For this, note that \[
f'(x)>0\iff \ln \frac{x+y}{x}>\frac{y}{x+y}=1-\frac{x}{x+y}.
\] Put $u= \frac{x+y}{x}$, the above inequality becomes $\ln u> 1-1/u$, where $u>1$, which can be proved either by the standard calculus or start from the convexity inequality $e^x\ge 1+x$ (then for $x>-1$, take reciprocal, take $\ln$, put $u=\frac{1}{1+x}$).
Example 4. \begin{align*}
\int_a^\infty \frac{n^2xe^{-n^2x^2}}{1+x^2}\,dx&=\int_{na}^\infty \frac{xe^{-x^2}}{1+x^2/n^2}\,dx \\
&\stackrel{\text{MCT}}{=\!=\!=} \int_{[na,\infty)}\frac{xe^{-x^2}}{1+x^2/n^2}\,dm\\
&= \int_{[0,\infty)}\chi_{[na,\infty)}\frac{xe^{-x^2}}{1+x^2/n^2}\,dm.
\end{align*} It is obvious that the integrant converges to $0$ pointwise, so it is enough to check the interchangeability of integral sign and limit. Observe that \[
\left|\chi_{[na,\infty)}\frac{xe^{-x^2}}{1+x^2/n^2}\right|\leq xe^{-x^2},
\] since $g(x)= xe^{-x^2}$ is Lebesgue integrable on $[0,\infty)$, so by LDCT we have \begin{align*}
\limn \int_a^\infty \frac{n^2xe^{-n^2x^2}}{1+x^2}\,dx&=\limn \int_{[0,\infty)}\chi_{[na,\infty)}\frac{xe^{-x^2}}{1+x^2/n^2}\,dm\\
&=\int_{[0,\infty)}\limn \chi_{[na,\infty)}\frac{xe^{-x^2}}{1+x^2/n^2}\,dm\\
&=\int_{[0,\infty)}0\,dm = 0.
\end{align*}
Example 5. We observe that \[
x\in E_k\implies x\in E_{k-1},
\] therefore $E_k\subseteq E_{k-1}$, i.e., $\{E_k\}$ is descending, we hope to use integration version of MST. Indeed, since $f$ is Lebesgue integrable, by descending version of MST (Monotone Set Theorem) we conclude \[
\lim_{k\to\infty}\int_{E_k}|f|\,dm = \int_{\capp_{k=1}^\infty E_k}|f|\,dm.
\] Since $f=0$ on $\capp_{k=1}^\infty E_k$, so \[
\lim_{k\to\infty}\int_{E_k}|f|\,dm=0.\]
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