# If $f_n,g_n \in L^1$ and $f_n,g_n \to 0$, show $\int_A (2f_n g_n)/(1+f_n^2+g_n^2)\to 0$, when $A$ has finite measure

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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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.
Hint: Egorov's theorem. Where there is uniform convergence, no problem, and on "the small set", the integrand is bounded by $1$.
The only suggestion you might still need is: $2f_ng_n \le f_n^2 + g_n^2$. This is what gives the bound ($\frac{a}{1+a} \le 1$).