# Let $f_n \ge 0$ converge pointwise to $f$ and any for simple function $\rho (x)\leq f(x)$ show that $\int\rho d\mu \leq \int f_n d\mu$

Let a sequence of nonnegative function $f_n$ on a measure space $(X,\beta, \mu)$ converge pointwise to $f$. Prove that for any nonnegative simple function $\rho (x)\leq f(x)$ we have $$\int\rho (x) d\mu \leq \liminf \int f_n (x)d\mu$$ assuming that the left hand side is finite.

I found this problem on a previous qualifying exam and I need help.

What I Think.
Before I proceed, let me ask this. Because $f_n$ is defined on a measure space $(X,\beta, \mu)$, can we say $f_n$ is measurable $\forall n$?

Presuming the above is true

If $f_n$ is measurable then the pointwise limit $f$ is also measurable. Since $f_n \ge 0$ and measurable, it is integrable and we invoke Fatou's lemma $$\int \liminf f_n d\mu \leq \liminf\int f_n d\mu \Rightarrow \int f d\mu \leq \liminf \int f_n d\mu \quad ........(1)$$

Let $\rho \ge 0$ be a simple function and $\rho (x)\leq f(x)$ , f is integrable [also nother presumption] and by definition $$\int f d\mu =sup\big\{\int \rho d \mu , \rho (x)\leq f(x)\big\}$$

Then from $(1)$ above, we have $$sup\big\{\int \rho d \mu , \rho (x)\leq f(x)\big\} \leq \liminf\int f_n d\mu$$ and hence $$\int \rho d \mu \le sup\int \rho d \mu \leq \liminf\int f_n d\mu$$

I believe I may have taken a lot of things for granted here. can someone correct me or write a proof it. thanks

This is a bit of a strange question, because if you're allowed to use Fatou's lemma then there's really no reason to consider only simple functions $\rho\leq f$. If you could find another way to prove the stated inequality, then it could be used to prove Fatou's lemma by taking the supremum over all simple functions $\rho\leq f$.
Anyway, to address your first concern, I think the problem implicitly assumes that the functions $f_n$ are measurable, for otherwise there's no way to make sense of $\int f_n\;d\mu$. If we can use Fatou's lemma, then we know that $$\int f\;d\mu\leq\liminf_{n\to\infty}\int f_n\;d\mu$$ and then if $g$ is any measurable function (not necessarily a simple function) with $0\leq g\leq f$ it follows that $$\int g\;d\mu\leq \int f\;d\mu\leq\liminf_{n\to\infty}\int f_n\;d\mu$$ by the monotonicity of the integral. Note that these inequalities hold even if some (or all) of these expressions are infinite.
• Okay, so if I understand you well, the question assumes that $f_n$ is measurable that is why it goes aheard to write the integral for it. But my question is , whenever I see the statement '$f_n$ on a measure space $(X, \beta, \mu$)' , does it mean that $f_n$ are all measurable w.r.t $\mu$? – J. Kyei Dec 29 '16 at 23:35