About a uniformly integrable sequence of functions

Question

I'm working with the following sequence of functions

$$\begin{cases} f(x), & f(x)\leq n, \\ n, & f(x)>n. \end{cases}$$ Here $f$ is a non-negative measurable function and $I=\lbrack 0,1\rbrack$. I'm told to prove the following:

1. Show that $\displaystyle\lim_{n\to\infty}\int_I f_n(x)\,dx=\int_{I}f(x) \, dx$.
2. Show that given $\varepsilon>0$, there exists a $\delta>0$ such that if $A\subseteq I$ with $m(A)<\delta$, then $\int_Af_ndx<\varepsilon$.
3. Now if we assume $f$ is integrable, show that if $A\subseteq I$ with $m(A)<\delta$, then $\int_Afdx<\varepsilon$.

I worked through this as follows and I would like to see if you can help me out with my mistakes and doubts.

I readily noticed that $f_n(x)=\min\{f(x),n\}$, however I don't see how that can help me. For part 1. I used the monotone convergence theorem. For this I (almost) proved that $(f_n)$ is an increasing sequence of functions. This is because

$$ f_n(x)\leq f_{n+1}(x)\iff \begin{cases} f(x) = f(x) &\text{ when } f(x)\leq n,\\ n\leq f(x) &\text{ when } n< f(x)\leq n+1,\\ n<n+1 &\text{ when } f(x)>n+1.\\ \end{cases} $$

The part of this argument I'm not so sure of is why does $n\leq f(x)$ when $ n< f(x)\leq n+1$?

I tried drawing the function and the sequence on a piece of paper and it does make sense that $f_{n+1}\geq f_n$ in any of those sets. My justification is along the lines of "if $f$ were less than $n$, it would occur that $f_n$ would also equal $f$ in that set and so $f=f$ gives no contradiction". I feel like my intuition is right but I can't formalize it.

To complete the first part and apply the MCT I have to show that there exists a function $\phi\in L(I)$ such that $\phi\leq f_n$ for all $n$. The prime candidate for this function would be $\phi=f_1$. The question then reduces to the integrability of $f_1$ which I worked like this

\begin{align*} \int_{I}f_1d x &=\int_{\{f>1\}}f_1d x+\int_{\{f\leq 1\}}f_1dx=\int_{\{f>1\}}1d x+\int_{\{f\leq 1\}}fdx\\ &=m(\{f>1\})+\int_{\{0\leq f\leq 1\}}fdx\leq m(\{f>1\})+m(\{0\leq f\leq 1\})\\ &=m(\{f\geq 0\}). \end{align*}

Now since $f$ is measurable the latter set is measurable. But now I have another problem, how do I handle the case when the measure of the set is infinite? Is my process of showing that $f_1$ integrable even correct?

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EDIT1: Since $\{f\geq 0\}\subseteq I$ and both are measurable sets, it follows that $m(\{f\geq 0\})\leq m(I)=1$. By this it follows that $f_1$'s integral is finite thus $f_1\in L(I)$. Now we can apply the MCT to show the first part. Is this reasoning correct?

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If it were the case that all the answers were positive then by MCT we get the first part of the problem.

For the second and third parts I'm stumped. I know that the condition in question is uniform integrability. I tried taking the integral of $f_n$ over $I$, but this time I separated it into $\{f>n\}$ and $\{f\leq n\}$ then it followed that

\begin{align*} \int_{I}f_nd x &=\int_{\{f>n\}}f_nd x+\int_{\{f\leq n\}}f_ndx=\int_{\{f>n\}}nd x+\int_{\{f\leq n\}}fdx\\ &=m(\{f>n\})+\int_{\{0\leq f\leq n\}}fdx\leq nm(\{f>n\})+nm(\{0\leq f\leq n\})\\ &=nm(\{f\geq 0\}). \end{align*}

How can I choose my $\delta$ with the quantity $nm(\{f\geq 0\})$? Once again I also have the question about my treatment of the past integral, is it done correctly?

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EDIT1: Following the same reasoning as the edit above, $nm(\{f\geq 0\})$ is bounded above by $n$. But this still doesn't lead me to a choice of an appropiate $\delta$...

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For the third part I can write $f$ as $f-f_n+f_n$ and since $f$ is integrable the difference of integrals has no problem. Then if $A$ is the set of the last item we have \begin{align*} \int_A|f| dx&\leq\int_A |f_n-f|dx+\int_A|f_n|dx\\ &<\eta m(A)+\varepsilon. \end{align*}

The last quantity is "small" so I feel like I can take $\eta=\varepsilon$ and then we will get a multiple of $\varepsilon$ so the whole integral is small. I feel like this needs to be formalized but I can't quite grasp it since I'm yet not sure about the second item.

This is the whole of my process, any comments and suggestions will be very welcome.

Answer 1

Question 1

Your approach looks fine. Note that when $ n < f(x) \le n+1$, $f_n(x) = n$ and $f_{n+1}(x) = f(x)$, so $f_n(x) \le f_{n+1}(x)$.

Showing that $f_1$ is integrable can also be done much more simply: since $\left\vert f_1 \right\vert \le 1$, $\int_I \left\vert f_1 \right\vert\,\mathrm d m \le m(I)$. You get to this eventually, but you needn't do so much work.

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Question 2

Uniform integrability is not needed here. Question $2$ only requires the stated property to hold for fixed $n$, so it is enough to observe that

$$ \int_A f_n \,\mathrm d m \le \int_A n \,\mathrm d m = nm(A). $$

Hence, we only need to choose $\delta = \frac{\varepsilon}{n}$.

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Question 3

This looks fine, but it is important to justify where you get your bound $\eta$. (Hint: use monotone/dominated convergence. You should also be able to do this in a way that doesn’t multiply $\eta$ by $m(A)$.)

Also, you can read about a different way to answer this question here.