Let ($\Omega, \mathcal{F}, \mu)$, with $\mu (\Omega) <\infty$ and suppose $f_n , n \geq 1$ is a sequence of integrable functions that converges uniformly on $\Omega$ to $f$. Given that $f$ is integrable, show that $$\int f_n\ d\mu \to \int f\ d\mu \quad\text{ as }\quad n \to \infty $$

I tried :

$\left|\int_\Omega f_n \, d\mu - \int_\Omega f\, d\mu\right| $

: (By linearity of the integral)

$= |\int_\Omega (f_n - f)\, d\mu|\leq \int_\Omega |f_n - f|\, d\mu$

How do we prove the inequality above?

I tried the following but I am not sure if it's correct or sufficient mathematical proof.

Since $f_n$ integrable, so $\int_\Omega f_n \ d\mu$ is finite which means it exists.

$f$ integrable, so$\int_\Omega f\ d\mu$ is finite, which means it also exists.

Thus, since they both exist $(f_n - f)$ exists $= |\int_\Omega (f_n - f)\, d\mu|$ is finite.

Hence from basic property of integral since $= |\int_\Omega (f_n - f)\, d\mu|$ exists

Finally how do we apply the Uniform convergence on $ \int_\Omega |f_n - f|\, d\mu$ to conclude that

$\int_\Omega f\ d\mu \to \int_\Omega f\ d\mu$ as $ n \to \infty $ ?

  • $\begingroup$ math.stackexchange.com/questions/2473228/… This is the same question but not fully explained . $\endgroup$
    – Mia
    Nov 9 '20 at 1:24
  • $\begingroup$ The first inequality follows from monotonicity of integrals. For the second question you must use the fact that $\int_{\Omega}\epsilon=\mu(\Omega)\epsilon$ $\endgroup$
    – 2132123
    Nov 9 '20 at 1:32
  • $\begingroup$ What should I choose as my $\epsilon $ ? $\epsilon$ + Μ(ω)? where M(ω) = max ${f_1(ω), ....f_{(N-1)}(ω)}$ ? Or use as $\epsilon$ the $\epsilon$ / $μ(\Omega)$ $\endgroup$
    – Mia
    Nov 9 '20 at 2:03

Since $f_n$ converges uniformly on $\Omega$ to $f$, for $\forall \varepsilon>0$, there is $N>0$ which is independent of $x\in\Omega$ such that $$ |f_n(x)-f(x)<\frac{\varepsilon}{\mu(\Omega)}, \forall n\ge N, x\in\Omega. $$ So for above $\varepsilon>0$ and $N>0$, when $n\ge N$, $$\bigg|\int_\Omega (f_n - f)\, d\mu\bigg|\le\int_\Omega |f_n-f|\ d\mu\le\int_{\Omega}\frac{\varepsilon}{\mu(\Omega)}d\mu=\varepsilon $$ which implies $$ \lim_{ n \to \infty}\int_\Omega f_n \, d\mu=\int_\Omega f_n \, d\mu. $$

  • $\begingroup$ Should we choose $|\Omega|$ or $|μ(\Omega)|$? Since $\int_{\Omega}d\mu$ = $μ(\Omega)$ $\endgroup$
    – Mia
    Nov 9 '20 at 2:13
  • $\begingroup$ They are the same. $\endgroup$
    – xpaul
    Nov 9 '20 at 2:17

Step 1. $$ \left|\,\int_\Omega f\,d\mu \,\right|\le \int_\Omega |f|\,d\mu $$ Proof. Set $$ f_-=\max\{-f,0\}, \quad f_+=\max\{f,0\}. $$ Then $f_-,f_+$ are measurable, $f_-,f_+\ge 0$ and also $f=f_+-f_-$ and $|f|=f_++f_-$. So $$ \left|\,\int_\Omega f\,d\mu \,\right|=\left|\,\int_\Omega (f_--f_-)\,d\mu \,\right| =\left|\,\int_\Omega f_+\,d\mu -\int_\Omega f_-\,d\mu \,\right| \le \left|\,\int_\Omega f_+\,d\mu\,\right| +\left|\,\int_\Omega f_-\,d\mu \,\right| \\=\int_\Omega f_+\,d\mu+\int_\Omega f_-\,d\mu =\int_\Omega (f_-+f_-)\,d\mu =\int_\Omega |f|\,d\mu. $$

Step 2. If $f_n\to f$ uniformly on $\Omega$, then for every $\varepsilon>0$, there exists a $N$, such that $$ n\ge N \quad\Longrightarrow\quad |f_n(x)-f(x)|<\frac{\varepsilon}{\mu(\Omega)+1} $$ for all $x\in\Omega$.

Hence, if $n\ge N$, then $$ \left|\,\int_\Omega f_n\,d\mu-\int_\Omega f\,d\mu\,\right|= \left|\,\int_\Omega (f_n-f)\,d\mu\,\right|\le \int_\Omega |f_n-f|\,d\mu \le \int_\Omega \frac{\varepsilon}{\mu(\Omega)+1}\,d\mu=\frac{\varepsilon\, \mu(\Omega)}{\mu(\Omega)+1}<\varepsilon. $$ and thus $$ \lim_{n\to\infty}\int_\Omega f_n\,d\mu=\int_\Omega f\,d\mu. $$


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