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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 $|f_n(x)| \le \phi(x)$.

Thanks in advance!

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$0\leq e^{-\frac{1}{n}\|x\|}\leq1$ 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.

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The easiest way (in my opinion) is not to use dominated convergence, but rather monotone convergence.

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    $\begingroup$ How exactly? We don't have $f \ge 0$. $\endgroup$ – mechanodroid Jul 16 at 10:32
  • $\begingroup$ @mechanodroid See my edited answer. $\endgroup$ – Feng Shao Jul 16 at 23:44

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