Convergence uniform of sequence of measurable non-negative functions in finite measure space.

Question

It's the first time I'm asking here. I'm having problem with an exercise on integration theory. It's:

Let $(X,\mathbb{X},\mu)$ a finite measure space. Denote $M^+(X,\mathbb{X}) = \{f \in [0,\infty]^X\:\:; f\:\: is\:\: measurable\}$. If $(f_n)$ is a sequence of measurable functions em $M^+(X,\mathbb{X})$ which converges uniformly to a function $f$, so $f \in M^+(X,\mathbb{X})$ and
\begin{equation} \int_Xfd\mu = \lim_{n \to \infty} \int_Xf_nd\mu \end{equation}.

Well, the natural manner of "solve" this is like it was done here Question 4.K of Bartle's Element of Integration, but i think it's incorrect, because this inequality just can be used if we can ensure that the function $|f_n-f|$ is integrable. I think we have to ask theses functions be integrable, or, since $\mu(X)<\infty$ and the convergence is uniform, the function $f$ is bounded.

Additional edition:

The function $f_n-f$ is, in fact, integrable for n enoughly large, but how can we ensure that \begin{equation} \int_Xf_nd\mu - \int_Xfd\mu = \int_X(f_n-f)d\mu \end{equation}

A priori, we just can ensure in two situations: Sum of non-negative measurable functions or sum/difference of integrable functions.

What do you think?

Answer 1

The functions $g_n=|f_n-f|$ are integrable eventualy since we are in a finite measure space.

Indeed by uniform convergence, for $\epsilon=1$ exists $N \in \Bbb{N}$ such that $g_n<1 ,\forall n \geq N,\forall x \in \Bbb{X}$

So $\forall n \geq N$ we have $\int g_n<\mu(X)<+\infty$

Now if $\int f=+\infty$ then the convergence is true by Fatou's Lemma.

If $\int f <+\infty$ then $\forall n \geq N$ then $$|f_n| \leq |f_n-f|+|f| \leq 1+f$$

So $f_n$ are eventualy integrable,so you have the conclusion