$ \lim _{n \rightarrow \infty} \int_{E} \frac{f_{n}^{2}(x)}{1+f_{n}^{2}(x)} \mathrm{d} m=0 $ associated with convergence in measure

Question

For $m E<+\infty$, why the sufficient and necessary condition of $\left\{f_{n}(x)\right\}$ converge in measure to $0$ is

$$ \lim _{n \rightarrow \infty} \int_{E} \frac{f_{n}^{2}(x)}{1+f_{n}^{2}(x)} \mathrm{d} m=0\;? $$

I have learnt that to show that $f_n\to 0$ in measure implies $\int_E\frac{|f_n|}{1+|f_n|}\to 0$, we should fix some $\varepsilon>0$, let $E_n=\{x:|f_n(x)|\geq \varepsilon\}$, and try splitting the integral over $E$ into an integral over $E_n$ and integral over $E\setminus E_n$. But I am not sure the exact ammount for $\varepsilon>0$ to fix and how should we split the $E_n$ and $E\setminus E_n$.

Anybody could help? Many Thanks.

Answer 1

Sufficiency - we need to argue that for every $\varepsilon > 0,$ $m(|f_n| \ge \varepsilon) \to 0$ as $n \to \infty$.

Fix any $\varepsilon,$ and define $E_n^{\varepsilon} = \{|f_n| \ge \varepsilon\},$ then notice that $$\int_E \frac{f_n^2}{1 + f_n^2} = \int_{E\setminus E_n^\varepsilon} \frac{f_n^2}{ 1 + f_n^2} + \int_{E_n^\varepsilon} \frac{f_n^2}{1 + f_n^2} \\ \ge \int_{E_n^\varepsilon} \frac{f_n^2}{1 + f_n^2} \ge m(E_n^\varepsilon) \min_{|f_n| \ge \varepsilon} \frac{f_n^2}{1 + f_n^2} = m(E_n^\varepsilon) \cdot C_{\varepsilon},$$ where $C_{\varepsilon} = \frac{\varepsilon^2}{1 + \varepsilon^2}$. Above the first inequality is because the integrand is non-negative, and the second is due to the monotonicity of integrals. Now if you send $n \to \infty,$ $\int_E (f_n^2/(1+f_n^2)) \to $, and thus $m(E_n^\varepsilon) C_\varepsilon \to 0$> But since $C_\varepsilon > 0,$ it has to be the case that $m(E_n^\varepsilon) = m(|f_n| \ge \varepsilon) \to 0$. Since $\varepsilon$ is arbitrary, this concludes the argument.

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For completeness, I'll also write the necessity argument. Notice that the sufficiency did not utilise the finiteness of $m(E),$ this only shows up in the necessity.

Define $E_n^\varepsilon$ and $C_\varepsilon$ in the same was as above. Since $f_n \to 0$ in measure, we know that for any fixed $\varepsilon,$ $m(E_n^\varepsilon) \to 0.$ Now, notice that $ \frac{f_n^2}{1 + f_n^2} \le 1,$ and that if $|f_n| \le \varepsilon,$ then $\frac{f_n^2}{1 + f_n^2} \le C_\varepsilon$. As a result, for every $\varepsilon > 0,$ it holds that

$$ \int_E \frac{f_n^2}{1 + {f_n^2}} = \int_{E \setminus E_n^\varepsilon} \frac{f_n^2}{1 + f_n^2} + \int_{E_n^\varepsilon} \frac{f_n^2}{1 + f_n^2} \\ \le m(E\setminus E_n^\varepsilon) C_\varepsilon + m(E_n^\varepsilon).$$ Sending $n \to \infty,$ we conclude that $$\forall \varepsilon > 0, \lim_{n \to \infty} \int_E \frac{f_n^2}{1 + f_n^2} \le m(E) C_\varepsilon. $$ But since this is true for all $\varepsilon > 0,$ and since the LHS does not depend on $\varepsilon$, we find that $$ \lim_{n\to 0} \int_E \frac{f_n^2}{1 + f_n^2} \le m(E) \lim_{\varepsilon \searrow 0} C_\varepsilon = 0,$$ where the final equality is due to the fact that $0\le C_\varepsilon \le \varepsilon^2.$ Of course, $\int_E \frac{f_n^2}{1 + f_n^2} \ge 0,$ and the conclusion follows.