Consider the measure space $(\mathbb{R},\mathscr{B} (\mathbb{R}),\lambda)$. Let $u \in L^1(\lambda)$ such that $ \int_{\mathbb{R}}^{} u \,d\lambda\ =3$.
I am trying to find out whether it is possible to find a sequence of functions $u_n \in L^1(\lambda) $ that converges to $u$ such that $ \int_{\mathbb{R}}^{} u_n \,d\lambda\ =1$ and $u_n \geq 0 $ for all $n\in\mathbb{N}$.
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My intuition tells me that the answer must be no. I have tried explaining it mathematically by Fatou's lemma.
If such a sequence $u_n$ is to exist, we know by Fatou's lemma that the following inequality holds:
$ \int_{\mathbb{}}^{} \lim_{n \to \infty} inf $ $ u_n $ $ \,d\lambda\ \leq \lim_{n \to \infty} inf \int_{\mathbb{}}^{} $ $ u_n $ $ \,d\lambda\ $
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But we find that: $\ $
$ \int_{\mathbb{}}^{} \lim_{n \to \infty} inf $ $ u_n $ $ \,d\lambda\ = \int_{\mathbb{}}^{} inf $ $ u $ $ \,d\lambda\ = \int_{\mathbb{}}^{} $ $ u $ $ \,d\lambda\ = 3 $
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And: $\ $
$ \lim_{n \to \infty} inf \int_{\mathbb{}}^{} $ $ u_n $ $ \,d\lambda\ = \lim_{n \to \infty} inf $ $ 1 =1 $
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Hence the inequality does not hold which tells us that such a sequence does not exist. Is my arguing here correct? And is the conclusion correct?