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I'm learning about measure theory, specifically the Lebesgue integral of nonnegative functions, and need help with the following problem:

Suppose that $f$ and $f_n$ are nonnegative measurable functions, that $f_n$ decreases pointwise to $f$, and that $\int_{\mathbb{R}}f_1 < \infty$. Prove that $\int_{\mathbb{R}}f = \lim\int_{\mathbb{R}}f_n$. [Hint: Consider $g_n = f_1 - f_n$]. Show with a counterexample that the assumption that $\int_{\mathbb{R}}f_1 < \infty$ is necessary.

The assumptions are (I always rewrite the problem to see if I understand it correctly):

  1. $\forall x \in \mathbb{R}, f_n(x) \geq f_{n+1}(x)$.
  2. $\int_{\mathbb{R}}f_1 < \infty$,that is $f_1 \in L^1(\mathbb{R})$.
  3. $f_n \to f$ pointwise in $\mathbb{R}$.

My work and thoughts:

As $f_n$ is monotone non-increasing, $g_n = f_1 - f_n$ is a monotone non-decreasing sequence of non-negative Lebesgue measurable functions, i.e $0 \leq g_1 \leq g_2 \leq \cdots \leq f_1 - f$ and $\lim g_n = f_1 - f$. Therefore, by the monotone convergence theorem

$$ \int_{\mathbb{R}}\lim_{n \to \infty} g_n = \int_{\mathbb{R}}(f_1 - f) = \lim_{n \to \infty}\int_{\mathbb{R}}(f_1 - f_n).$$

Because $\int_{\mathbb{R}}f_1 < \infty$ and $f_n \leq f_1$ for all $n \in \mathbb{N}$, this implies that $\int_{\mathbb{R}}f_n < \infty$ for all $n \in \mathbb{N}$.


This is where I'm stuck. I think I'm really close to the desired result. How do I continue from here? Also how do I show that the assumption that $\int_{\mathbb{R}}f_1 < \infty$ is necessary using a counterexample?

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$$ \int_{\mathbb{R}}f_1 - \int_{\mathbb{R}}f =\int_{\mathbb{R}}(f_1 - f) =\int_{\mathbb{R}}\lim_{n \to \infty} g_n = \lim_{n \to \infty}\int_{\mathbb{R}}(f_1 - f_n) = \int_{\mathbb{R}}f_1 - \lim_{n \to \infty}\int_{\mathbb{R}}f_n$$

Subtract $\int_{\mathbb{R}}f_1$ from both sides and you are done.

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    $\begingroup$ @VonKar You can subtract $\int_{\mathbb{R}}f_1$ from both sides, because $\int_{\mathbb{R}}f_1<\infty$. $\endgroup$ – Ramiro Apr 18 '16 at 2:40

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