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Question:

$(X,\mathcal{M},\mu)$ measure space. Suppose that $\{f_n\}\subset L^1$ and $f_n\rightarrow f$ uniformly. Show that if $\mu(X) <\infty$, then $\int f_n\rightarrow \int f$.

Proof:

I exploit uniform convergence and $\{f_n\}\subset L^1$ to say that $\forall x$ $\exists N$ s.t. $n\geq N \Rightarrow |f_n|\leq M$ where M is finite, then I build the dominating function $g:=M \chi_{[X]}$ and apply the DCT to the sequence $\{f_n\}_{n\geq N}$ and then claim that $\lim_{n\geq N\rightarrow\infty}\int f_n=\lim_{n\rightarrow\infty}\int f_n$.

Makes sense? (I am new to math, please show mercy)

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2 Answers 2

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Not quite: The functions $\frac{1}{\sqrt{x}} - \frac{1}{n}$ converge uniformly to $\frac{1}{\sqrt{x}}$ on $(0, 1)$ even though none of these functions is bounded. More generally, there's no reason to expect an $L^1$ function to be bounded.

What you can say, however, is that there exists an $N$ such that whenever $n \ge N$, $|f(x) - f_n(x)| < \epsilon$ for all $x$ in the space. Then estimate

$$\left|\int f_n - \int f\right| \le \int |f_n - f| \le \int \epsilon = \epsilon \mu(X) $$


Alternatively, using uniform continuity, you can show that (eventually) the function $$g(x) = |f(x)| + 1$$ dominates your sequence.

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  • $\begingroup$ Very clear, thanks for prompt and detailed reply. $\endgroup$
    – Rgkpdx
    Dec 1, 2013 at 10:15
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    $\begingroup$ @tbongers To use the inequality $|\int f_n - \int f| \le \int |f_n -f|$ you don't have to prove first that $f \in L^{1}$? Thanks! :) $\endgroup$
    – shamisen
    May 17, 2014 at 15:42
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    $\begingroup$ How we can write | ∫fn− ∫f | ≤ ∫| fn−f | without proving f is integrable? $\endgroup$
    – Sushil
    Nov 25, 2014 at 11:25
  • $\begingroup$ In the second approach, it should be also proved that $f \in L^1$ so that $g \in L^1$. $\endgroup$ Oct 9, 2016 at 16:19
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Here is the complete proof when $f_j: X \to \mathbb{R}$ is continuous and $\mathcal{M}$ is the Borel $\sigma$-algebra $\mathcal{B}_X$. By the uniform convergence $f_j \to f$ and the uniform limit theorem, $f$ is continuous and thus measurable by $\mathcal{M} = \mathcal{B}_X$. By the uniform convergence $f_j \to f$ again, $\forall \varepsilon > 0$: $\exists N \in \mathbb{N}$ such that \begin{equation} |f_n(x) - f(x)| < \varepsilon \;\;\, \forall x \in X \;\;\, \forall n \geq N, \end{equation} by which we obtain for any $n \geq N$: \begin{equation} \int_X \big | \, f_n(x) - f(x) \big | \, d\mu(x) \leq \int_X \varepsilon \, d\mu(x) = \varepsilon \cdot \mu(X), \end{equation} where "$f_n - f$" is measurable and, by $\mu(X) < \infty$, belongs to $L^1$. Since $f_n \in L^1$ and $f - f_n \in L^1$, we have $f \in L^1$. Hence $f_j \to f$ in $L^1$, which completes the proof.


[Note] By $f_n$, $f \in L^1$ and the above integral inequality, it is now obvious that \begin{equation} \bigg | \int_X \, f_n(x)\; d\mu(x) - \int_X \, f(x)\; d\mu(x) \bigg | \leq \int_X \big | \, f_n(x) - f(x) \big | \, d\mu(x) \leq \varepsilon \cdot \mu(X) < \infty. \end{equation}

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