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Problem Statement

Let $f$ be a non-negative measurable function. Prove that $$\lim_{n \to \infty} \int \min(f,n) \rightarrow \int f$$.

Attempt

First, if $f= \infty$ on a set of positive measure, $E$, then $\int f\geq\int_E f=\infty$ so $\int f=\infty$. On the other side, \begin{eqnarray}\int \min(f,n) &= \int_E n + \int_{E^c} \min(f,n) \\ &\geq n\mu(E) \\ &\rightarrow \infty,\end{eqnarray} since $f \geq 0$ and $\mu(E)>0$.

Next, suppose that $f= \infty$ on a set of $0$ measure. In this case, let $A_n=\{x:n<f\}$ so that $A_{n+1} \subset A_n$ and \begin{eqnarray}\int\min(f,n)&=\int_{A_n^c} f + \int_{A_n} n \\ &\int_{A_n^c} f + n\mu(A_n).\end{eqnarray} Now $A_n$ is a descending sequence of sets approaching $\varnothing$ and if $x \in X$, then for some $n$, $x \not\in A_i$ for all $i>n$ so $x \in A_i^c$ for all $i>n$. Thus $A_i^c$ is an ascending sequence of sets approaching $X$.

Question

Is this correct?

Edit

Convergence theorems are not known at this point in the text

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    $\begingroup$ $\mathrm{min}(f,n)$ is a non negative increasing sequence and tends to $f$ as $n$ tends to $+\infty$ : you can apply Beppo-Levi theorem. $\endgroup$ – Nicolas Jun 15 '15 at 17:34
  • $\begingroup$ I should add that I have no convergence theorems to work with at this point in the text $\endgroup$ – illysial Jun 15 '15 at 17:35
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First notice that for each $n$: $$\min(f,n)\leq\min(f,n+1)\leq f$$ hence: $$\int\min(f,n)\leq\int\min(f,n+1)\leq\int f$$ This proves that it makes sense to speak about $\lim_{n\rightarrow\infty}\int\min(f,n)$ and that:$$\lim_{n\rightarrow\infty}\int\min(f,n)\leq\int f$$

Let $g\in\mathcal G$ if it is a non-negative measurable function that takes only a finite number of values and satisfies $g(x)\leq f(x)$ for each $x$. Then function $g$ has a maximum value and if $n$ exceeds this maximum value then $g(x)\leq \min(f(x),n)$ for each $x$.

This implies: $$\int g\leq\lim_{n\rightarrow\infty}\int\min(f,n)$$ and consequently: $$\int f:=\sup\{\int g\mid g\in\mathcal G\}\leq\lim_{n\rightarrow\infty}\int\min(f,n)$$

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Your argument is incomplete.

  • Why are we "done" in the case where $f = \infty$ on a set of positive measure? You should argue this carefully.

  • In your last line, you need to explain why $A_n^c \to X$ (by which what you really mean is $A_1^c \subset A_2^c \subset \dots$ and $\bigcup_n A_n^c = X$) implies $\int_{A_n^c} f \to \int f$. (If you know the monotone convergence theorem (MCT), you could use it, but see below...)

  • You also have not said anything about the $n \mu(A_n)$ term; you seem to be assuming that it approaches 0. Since you know $A_n \to \emptyset$ (by which you mean $A_1 \supset A_2 \supset \dots$ and $\bigcap_n A_n = \emptyset$), you can use continuity from above to conclude $\mu(A_n) \to 0$, but that is not sufficient; what if we had something like $\mu(A_n) = n^{-1/2}$? In fact, it is possible that this limit is not zero, though it turns out that this only happens in the case when $\int f = \infty$; but as you can see, this part is harder than it looks.

If, as you say, you are not allowed to use any convergence theorems, then all you really have to work with is the definition of the integral, via simple functions. I would start over with a different approach, based on the following hint.

Suppose first that $\int f < \infty$. Fix $\epsilon > 0$. By definition of the integral, there is a simple function $g \le f$ such that $\int g > \int f - \epsilon$. But simple functions are bounded - there is an $n$ such that $g \le n$...

If instead we have $\int f = \infty$, for any $M > 0$ there is a simple function $g$ with $g \le f$ and $\int g \ge M$. Again there is $n$ such that $g \le n$...

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  • $\begingroup$ Ah...I think I can work with your approach...however, is there any way to deal with your third bullet point? I think I dealt with the first two points..maybe my method was fruitless..I will update my question using your method instead. $\endgroup$ – illysial Jun 15 '15 at 18:11
  • $\begingroup$ I would usually do it with the dominated convergence theorem. Presumably one can do it from scratch, though I would have to think about it. $\endgroup$ – Nate Eldredge Jun 15 '15 at 18:13

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