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Let $(f_n)$ and $(g_n)$ be sequences of nonnegative function in $L^1(\mathbb R)$, for which $$ f_n \to 0, \\ g_n \to 0, $$ almost everywhere. Show $$ \int_A \frac{2f_n g_n}{1+f_n^2+g_n^2} \to 0, $$ when $A \subset \mathbb R$ is a set of finite measure.

Define $$ h_n = \frac{2f_n g_n}{1+f_n^2+g_n^2}. $$ I know $$ h_n \le 2f_n g_n, $$ but I can't use a convergence theorem, because the product of integrable function may fail to be integrable.
I considered uniform integrability $$ \text{Since } f_n \in L^1: \forall \epsilon, \exists \delta : m(B) < \delta \to \int_B f_n <\epsilon, $$ where $m(\cdot)$ is Lebesgue measure.
To use that, I write $A=\cup_m B_{n,m}$ where the index $n$ is associated to $f_n$.
I can't go further.

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3 Answers 3

up vote 3 down vote accepted

You have $0 \le h_n \le g_n$ since $\dfrac{2f_n}{1 + f_n^2} \le 1$. Thus $h_n \to 0$. You also have $0 \le h_n \le 1$ since $\dfrac{2f_n g_n}{1 + f_n^2 + g_n^2} \le 1$. Since $A$ has finite measure, this provides a majorant of the sequence.

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Hint: Egorov's theorem. Where there is uniform convergence, no problem, and on "the small set", the integrand is bounded by $1$.

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The only suggestion you might still need is: $2f_ng_n \le f_n^2 + g_n^2$. This is what gives the bound ($\frac{a}{1+a} \le 1$).

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