# Application of Dominated Convergence

I need help with this problem:

Suppose that $$f$$ is a Lebesgue integrable function defined on $$\mathbb{R}^d$$. Prove that $$\int_{\mathbb{R}^d} f(x)e^{-\frac{1}{n}\|x\|}dx \to \int_{\mathbb{R}^d} f(x)dx.$$

If I define $$f_n(x) = f(x)e^{-\frac{1}{n}||x||}$$, then I am having trouble finding a suitable integrable $$\phi(x)$$ such that $$|f_n(x)| \le \phi(x)$$.

$$0\leq e^{-\frac{1}{n}\|x\|}\leq1$$ so that $$|f_n|=|f|e^{-\frac{1}{n}\|x\|}\leq |f|$$ and then you can use DCT.
Edit: We can also use MCT: First we have $$\left|\int_{\mathbb R^d}f(x)e^{-\frac{1}{n}\|x\|}\,dx -\int_{\mathbb{R}^d} f(x)\,dx\right|\leq\int_{\mathbb{R}^d}|f(x)|(1-e^{-\frac{1}{n}\|x\|})\,dx.$$ Since $$|f(x)|e^{-\frac{1}{n}\|x\|}\nearrow|f(x)|$$ for all $$x\in\mathbb R^d$$, an application of MCT to $$|f(x)|e^{-\frac{1}{n}\|x\|}$$ gives that $$\int_{\mathbb R^d}|f(x)|e^{-\frac{1}{n}\|x\|}\,dx\nearrow\int_{\mathbb R^d}|f(x)|\,dx$$ and therefore $$\int_{\mathbb{R}^d}|f(x)|(1-e^{-\frac{1}{n}\|x\|})\,dx\to0\ \ \text{as } n\to\infty,$$ so $$\int_{\mathbb R^d}f(x)e^{-\frac{1}{n}\|x\|}\,dx \to\int_{\mathbb{R}^d} f(x)\,dx\ \ \text{as } n\to\infty.$$ But unlike the other answer, my opinion is that it is easier to use DCT than MCT.
• How exactly? We don't have $f \ge 0$. – mechanodroid Jul 16 at 10:32