# If $f_n$ converges uniformly to $f$, then $\int f_n\ d\mu \to \int f\ d\mu$ as $n \to \infty$

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 $$n \to \infty$$ ?

• math.stackexchange.com/questions/2473228/… This is the same question but not fully explained .
– Mia
Nov 9 '20 at 1:24
• The first inequality follows from monotonicity of integrals. For the second question you must use the fact that $\int_{\Omega}\epsilon=\mu(\Omega)\epsilon$ Nov 9 '20 at 1:32
• 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 $\epsilon$ / $μ(\Omega)$
– 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 $$\lim_{ n \to \infty}\int_\Omega f_n \, d\mu=\int_\Omega f_n \, d\mu.$$
• Should we choose $|\Omega|$ or $|μ(\Omega)|$? Since $\int_{\Omega}d\mu$ = $μ(\Omega)$
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 $$\lim_{n\to\infty}\int_\Omega f_n\,d\mu=\int_\Omega f\,d\mu.$$