I had learned the usual dominated convergence theorem which says that if $f_n \to f$, and each $f_n$ bounded by some integrable g, then $\int f_n \to \int f$. However, I came across a "different" version in one of the examples, where dominated convergence was used to justify that if x tends to $x_o$ implies f(x) tends to $f(x_0)$ then $\int f(x) \to \int f(x_0)$. Could someone show why (presumably) the original one implies this one? Thanks!
Lebesgue Dominated Convergence theorem where $x \to x_0$ instead of $ n \to \infty$?
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real-analysis
lebesgue-integral
lebesgue-measure
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1For your second example, please put in some $dx$ or $d\mu$ so we can tell what you are talking about. As it stands, I cannot tell what you mean. – 2017-01-06
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0I agree with GEdgar. However, to interpret to my best guess, it looks like $x$ now represents a generalized index variable (generalizing "$n$"), rather than an integration variable. So, for example, $\int f_x(t) dt \rightarrow \int f(t) dt$ as $x \rightarrow x_0$, or $\int f_x d\mu \rightarrow \int f d\mu$ as $x \rightarrow x_0$. – 2017-01-06
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0The reason this works is just because, for any function $h(y)$, we have $\lim_{y\rightarrow y_0} h(y)=c$ if and only if $\lim_{n\rightarrow\infty} h(y_n)=c$ for every sequence of points $\{y_n\}_{n=1}^{\infty}$ that satisfies $y_n \rightarrow y_0$ (and $y_n \neq y_0$ for all $n$). – 2017-01-07
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0Hi, the actual example was: note that for every x, $f(x)e^{−2πix·ξ} → f(x)e ^{−2πix·ξ_0} $ as $ξ → ξ_0$, where $ξ_0$is any point in R, hence $\hat{f(ξ)} → \hat{f(ξ_0)}$ by the dominated convergence theorem. – 2017-01-07
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3Yes, in that case $x$ is the integration variable while $\xi$ is the index variable. So you have $\int g_{\xi_n}(x)dx \rightarrow \int g_{\xi_0}(x)dx$ for any sequence $\{\xi_n\}_{n=1}^{\infty}$ that satisfies $\xi_n\rightarrow \xi_0$. You can define $r_n(x) = g_{\xi_n}(x)$ if you like. – 2017-01-07
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0I think I get it, I will look at it more closely when I get home. Thanks! – 2017-01-07