Monotone Convergence Theorem for Riemann Integrable functions

I'm having a really hard time proving this statement (this is not homework):

If $f_{n} : [0,1] \rightarrow \mathbb{R}$ is a Riemann integrable function for all $n \in \mathbb{N}$, and $0 \leq f_{n + 1} \leq f_{n}$, and $\lim \limits_{n \rightarrow \infty} f_{n} = 0$, I need to prove that $\lim \limits_{n \rightarrow \infty} \int \limits_{0}^{1} f_{n}(x) \text{ } \mathrm{d}x = 0$.

I'm not allowed to use the Monotone Convergence Theorem for Riemann integrable functions (proving this is actually the first step in proving the MCT).

Now, I know $\lim \limits_{n \rightarrow \infty} \int \limits_{0}^{1} f_{n}(x) \text{ } \mathrm{d}x$ exists because the sequence $\left \{ \int \limits_{0}^{1} f_{n}(x) \text{ } \mathrm{d}x \right \}_{n =1}^{\infty}$ is a monotonically decreasing sequence that is bounded from below by $0$. However, I have no idea how to prove the limit is $0$.

Also, there is a hint to the problem. Assume $\lim \limits_{n \rightarrow \infty} \int \limits_{0}^{1} f_{n}(x) \text{ } \mathrm{d}x = \epsilon > 0$. I must choose a partition $P_{n}$ for $f_{n}$ such that $P_{n + 1}$ is a refinement partition of $P_{n}$ and show that there exists an element in $[0,1]$ such that $f_{n}$ converges to some strictly positive value on that element, which would lead to a contradiction of the hypothesis of pointwise convergence to $0$.

I thought of using a sequence of closed intervals that are nested, because their intersection would be nonempty (since closed subsets of compact spaces are compact), but I can't construct the sequence. If the hint makes the problem harder, is there an easier way to prove this statement? Any help would be greatly appreciated.

• Here's one proof (not using the hint). – David Mitra Jul 21 '14 at 18:28
• @David Mitra Thanks, I found that on Google when I was searching. This is a hard problem! It doesn't seem like it should be, though. – layman Jul 21 '14 at 20:10
• Hard problem? Yes. Back when I learned about Lebesgue integration, the instructor gave this as an example of the superiority of the Lebesgue theory over the Riemann theory. – GEdgar Mar 3 '15 at 15:50
• Here is another proof by C. Niculescu and F. Popovici (2011), theorem 2. It doesn't use the hint but Cousin lemma and Lebesgue criterion of Riemann integrability. Put $A=\varnothing$ in the proof. – Tony Piccolo Nov 15 '15 at 8:45
• @Tony Piccolo Here is another to add to the collection. It appeared in the Monthly 2010: classicalrealanalysis.info/documents/… – B. S. Thomson Nov 17 '15 at 20:49

We need to prove the following lemma:

Lemma: If $$f$$ is non-negative and Riemann integrable on $$[a, b]$$ with $$\int_{a}^{b}f(x)\,dx > 0$$ then there is a sub-interval of $$[a, b]$$ on which $$f$$ is positive.

The lemma is proved easily by using the definition of Riemann integral. Let $$I = \int_{a}^{b}f(x)\,dx > 0$$ and let $$0 \leq f(x) < M$$ for all $$x \in [a, b]$$. By definition of Riemann integral there is a partition $$P = \{x_{0}, x_{1}, x_{2}, \dots, x_{n}\}$$ of $$[a, b]$$ such that the Riemann sum $$S(f, P) = \sum_{i = 1}^{n}f(t_{i})(x_{i} - x_{i - 1}) > \frac{I}{2}$$ Let $$\epsilon = \frac{I}{2(M + b - a)}$$ and consider the set $$J = \{x: x \in [a,b], f(x) \geq \epsilon\}$$. Now we split the Riemann sum $$S(f, P)$$ as $$S(f, P) = \sum_{i \in A}f(t_{i})(x_{i}- x_{i - 1}) + \sum_{i \in B}f(t_{i})(x_{i}- x_{i - 1})$$ where $$A = \{i: [x_{i - 1}, x_{i}] \subseteq J\}, B = \{i: i \notin A\}$$ Now for $$i \in A$$ we can see that $$f(t_{i}) \leq M$$ and for $$i \in B$$ we can choose $$t_{i}$$ such that $$f(t_{i}) < \epsilon$$. Then we can see that $$\epsilon(M + b - a) = \frac{I}{2} < S(f, P) < M\sum_{i \in A}(x_{i} - x_{i - 1}) + \epsilon(b - a)$$ It now follows by simple algebra that $$\sum_{i\in A}(x_{i} - x_{i - 1}) > \epsilon$$ and therefore there exist intervals of type $$[x_{i - 1}, x_{i}]$$ where $$i \in A$$ where $$f(x) \geq \epsilon$$. The above proof is taken from my favorite book Mathematical Analysis by Tom M. Apostol.

Now we need to make use of this lemma for solving the current problem. Let's assume that $$\lim_{n \to \infty}\int_{0}^{1}f_{n}(x)\,dx = \lim_{n \to \infty}I_{n} = c > 0$$ Then it follows that after a certain value of $$n$$ we have $$I_{n} > c/2$$. And by our lemma it follows that $$f_{n}(x) \geq d > 0$$ where $$d$$ is some constant dependent only on $$c$$ and not on $$n$$. Moreover we see that there is a partition $$P_{n}$$ of $$[0, 1]$$ such that $$f_{n}(x)\geq d$$ in some of the sub-intervals made by $$P_{n}$$ (whose total length also exceeds $$d$$) and $$f_{n}(x) < d$$ for some point in remaining sub-intervals. If necessary we can replace $$P_{n + 1}$$ by $$P_{n} \cup P_{n + 1}$$ and hence it is OK to assume that $$P_{n + 1}$$ is a refinement of $$P_{n}$$.

Since $$f_{n}$$ is monotonically decreasing it follows that the sub-intervals of partition $$P_{n + 1}$$ where $$f_{n + 1}(x) \geq d$$ are contained in the corresponding sub-intervals of $$P_{n}$$ and again these sub-intervals of $$P_{n + 1}$$ have a total length greater than $$d$$. Thus it follows that there is a sub-interval of $$[0, 1]$$ where $$f_{n}(x) \geq d > 0$$ for all $$n$$ after a certain value. This contradicts the fact that $$f_{n}$$ converges pointwise to $$0$$.

The above proof is wrong and the wrong inferences are striked out. The approach however can be salvaged with some more effort. An alternative approach is given in this answer.

The following proof uses the concept of uniform convergence.

First note that $$\;f_n \to 0$$ on $$[0,1]$$, which is a compact set, the limit function is continuous and $$\;f_n$$ decreases monotonically. Then $$\;f_n$$ converges uniformly to $$0$$ by Dini's theorem. Thus $$\begin{equation} \lim_{n \to \infty} \int_0^1 f_n(x) \operatorname{d}\!x = \int_0^1\lim_{n \to \infty} f_n(x) \operatorname{d}\!x = \int_0^1 0 \operatorname{d}\!x = 0. \end{equation}$$ (The interchange of limits is legitimate due to uniform convergence of $$f_n$$ and the fact that the interval $$[0,1]$$ is compact.)

Since $\int_0^1 f_n$ is monotone decreasing in $n$, argue by contradiction, we assume that $$\int_0^1 f_n \geq 2 \epsilon \quad \forall n.$$ By the definition of Riemann integral $\int_0^1 f_n = \sup_{g\leq f_n, \text{ piecewise constant }} \int_0^1 g$, there exists a piecewise constant function $g_n$ for each $f_n$ such that $$\int_0^1 g_n \geq \epsilon\quad \forall n.$$ Now, $g_n \leq f_1$ for each $n$, and because $f_1$ is Riemann integrable, $f_1$ is bounded above by a constant $C$ , thus we have $g_n\leq C$. $$\epsilon \leq \int_0^1 g_n = \sum a^n_i |I^n_i| \leq C \sum |I^n_i|$$ which means that $$\frac{\epsilon}{C}\leq \sum |I^n_i| \quad \forall n.$$ The above inequality implies that the sum of the length of the intervals which each $g_n$ takes a positive value is bounded below by $\frac{\epsilon}{C}$ for each $n$. And such $g_n\leq f_n$ exists for all $f_n$ , this contradicts the fact that $f_n \rightarrow 0$ pointwise.

Edit: If you are allowed to use results from measure theory and Lebesgue integration, see Bounded Convergence theorem or Dominated Convergence theorem.

• I don't get the end. These intervals might not be nested as $n$ increases. – Ted Shifrin Jul 21 '14 at 23:07
• The bound $\frac{\epsilon}{C}$ does not depend on $n$. So for each $f_n$ there exists a function $g\leq f_n$ such that $m(\{x: g(x) > 0\}) \geq \frac{\epsilon}{C}$. And because the domain is bounded, the mass can not escape like the function $\chi_{[n,n+1]}$ define on $\mathbb{R}$. So you will not get $f_n \rightarrow 0$ pointwise. – Xiao Jul 22 '14 at 7:47
• Please explain: you say "there exists a piecewise constant function $g$ for each $f_n$ such that ..." So the choice of $g$ depends on $f_n$ in which case there should be a label $g_n$ that depends on $n$. And then how does the proof go? – B. S. Thomson Nov 17 '15 at 22:26
• I didn't notice the date so Mr Xiao is probably not around to answer, so I leave it as a fatal flaw in an attempted proof. It is not quite that easy, although maybe someone can find easier proofs than those published. – B. S. Thomson Nov 18 '15 at 0:32
• @B.S.Thomson I kinda just did a quick edit of the answer as you said. I will take a closer look of this. – Xiao Nov 18 '15 at 0:41

The following proof uses the countable additivity of Jordan content on its domain (i.e. if one knows from the start that the disjoint union has Jordan content).

Let $A_i=\{(x,y) \in [0,1] \times \mathbb R : 0<y<f_i(x)\}$.

By the monotone convergence of $\{f_i\}$ to zero, $A_{i+1} \subseteq A_i$ and $\,\bigcap_{i=1}^ \infty A_i = \varnothing$.

So, if $\,B_i=A_i \setminus A_{i+1}$, then $\{B_i\}$ is a disjoint sequence and $\,\bigcup_{i=1}^ \infty B_i = A_1$.

By the R-integrability of $f_i$, the set $\,B_i\,$ has Jordan content (in $\mathbb R^2$) and , by the countable additivity of Jordan content on its domain, the series $\sum_{i=1}^ \infty m(B_i)$ is convergent, so the sequence of its remainders $\{\sum_{i=n}^ \infty m(B_i)\}$ is null.

But $\,\sum_{i=n}^ \infty m(B_i) = m(A_n)= \int_0^1 f_n \,\dots$