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I try to solve the following task:

Let $(\Omega,\mathfrak{A},\mu)$ be a measurable space and $\mu(\Omega)<\infty$. Let $(f_n)_{n\geq1}$ be a sequence of integrable measurable functions $f_n:\Omega \rightarrow [-\infty,\infty]$ converging uniformly on $\Omega$ to a function $f$. Prove that $$\int f d\mu = \lim\limits_{n\rightarrow \infty} \int f_n d\mu$$.

What I thought: uniformly convergence of $f_n \rightarrow f$ means that $f$ is continuous and therefore measurable. Now I thought that I could use the dominated convergence theorem to show the equality.

uniformly convergence means $\lim\limits_{n\rightarrow \infty} \sup \{|f_n-f(x)|:x\in \Omega\}=0$ so I think I can define a function $s(x):=\sup \{|f_n-f(x)|:x\in \Omega\}=0$ which dominates all the $f_n$ and apply the theorem. But I'm not sure, if this is the correct way.

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    Uniform convergence does not imply continuity (under what topology on $\Omega$ ?) unless you have the additional hypothesis that the $f_n$ are continuous. – Justpassingby Nov 26 '15 at 12:40

2 Answers2

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Fix an $\varepsilon > 0$, by uniform convergence, we know that there exists $N \in \mathbb{N}$ such that for $n \geq N$, \begin{equation*} |f_n| = |f_n - f + f| \leq |f_n - f| + |f| < |f| + \varepsilon \end{equation*} Then define the function $g : \Omega \to \mathbb{R}$ by $g(\omega) = |f(\omega)| + \varepsilon$. Then $g$ is integrable, since we work on a finite measure space. If it was an infinite measure space, the $"+ \varepsilon"$ part would give some difficulties. Next, define $h_n = f_{N + n}$, with $\lim\limits_{n \to \infty} h_n = f$, for which it holds that $|h_n| \leq g$. Then, all conditions of the dominated convergence theorem are satisfied, hence we can conclude \begin{equation*} \lim_{n \to \infty} \int_\Omega h_n \mathrm{d}\mu= \int_\Omega f \mathrm{d}\mu \tag{$\ast$} \end{equation*} Lastly, observe that the difference between $\{f_n\}$ and $\{ h_n \}$ is only an shift in indices, hence it immediately follows from ($\ast$) that \begin{equation*} \lim_{n \to \infty} \int_\Omega f_n \mathrm{d}\mu = \lim_{n \to \infty} \int_\Omega h_n \mathrm{d}\mu= \int_\Omega f \mathrm{d}\mu \end{equation*}

Nigel Overmars
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    Why $g$ is integrable? – Rafael Jan 24 '16 at 21:24
  • @RafaelHolanda I know it's been 5 years but perhaps for anyone coming later, I will explain it. $f$ being integrable implies $|f|$ is integrable, which makes $g$ integrable.

    P/S: In an earlier comment which I deleted, I confused integrability with the existence of the integral of $f$ which is simply implied by measurability.

     – Duc Nguyen Oct 14 '21 at 08:21
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Uniform convergence and measurability of fn imply that f is measurable. Then |f| is measurable. We claim that f is integrable. Indeed, by uniform convergence, ∃ N s.t ∫|f| ≤ ∫ (|f - fn| + |fn|) ≤ ∫|f - fn| + ∫|fn| < ∞ for all n≥N, so f is integrable. Then by similar argument, ∫f - ∫fn = ∫ (f - fn) ≤ ∫|f - fn| ≤ ε·μ(Ω). Hence, the equality follows.

Deepleeqe
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