Study the convergence of the sequence of functions $$f_n(x)= \frac{f(x)}{1+\frac{|f(x)|}{n}}$$ (convergence in measure, pointwise and in $ L^2(\mathbb{R} ^d)$).
Let f be a measurable function such that $f \in L^2(\mathbb{R} ^d)$
$$f_n(x)= \frac{f(x)}{1+\frac{|f(x)|}{n}}$$
My attempt:
1) $\{f_n\} \to f(x)$ pointwise on $\mathbb{R}^d$
2)Convergence in measure:
for all $\lambda>0$, by the Chebyshev's inequality:
$$\lim_{n \to +\infty} m \left( \left\{ x \in \mathbb{R}^d : |f_n-f|=\left|\frac{f(x)^2}{n+|f(x)|}\right|> \lambda \right\} \right)<\lim_{n\to +\infty} \int_{\mathbb{R}^d}\frac{f(x)^2}{n+|f(x)|} $$
Now:
$$g_n(x)= \frac{f(x)^2}{n+|f(x)|} \to g(x)=0$$ pointwise on $\mathbb{R}^d$ and $f(x)^2$ dominates $g_n$ on $E$ in the sense that $|g_n|<f^2$ on $\mathbb{R}^d$ for all $n$.
Therefore, by the lebesgue dominated convergence theorem:
$$\lim_{n\to +\infty} \int_{\mathbb{R}^d}\frac{f(x)^2}{n+|f(x)|} = \int_{\mathbb{R}^d} \lim_{n\to +\infty}\frac{f(x)^2}{n+|f(x)|}=0 \\ \implies \lim_{n \to +\infty} m(\{x \in \mathbb{R}^d : |f_n-f|> \lambda\})=0$$
I would be thankful if somebody tell me whether my attempt is correct or not :)
Yes you are correct. Also, $L^{2}$ convergence follows similarly. Note that
\begin{eqnarray*} \left\Vert f_{n}-f\right\Vert _{2}^{2} & = & \int\frac{f(x)^{4}}{\left(n+\left|f(x)\right|\right)^{2}}. \end{eqnarray*} But since $\frac{f(x)^{4}}{\left(n+\left|f(x)\right|\right)^{2}}\to0$ pointwise and since $$ \frac{f(x)^{4}}{\left(n+\left|f(x)\right|\right)^{2}}\leq f(x)^{2}\in L^{1} $$ then again by dominated convergence we get that $\left\Vert f_{n}-f\right\Vert _{2}^{2}\to0$.