Let $h:[0,1] \times [0,1] \rightarrow \mathbb{R}$ be the function $h(x,y)=f(x)g(y)$. Show h is integrable.

Question

Let $f,g:[0,1] \rightarrow R$ be bounded, nonnegative, and nondecreasing $f(x_1) \leq f(x_2)$ for all $x_1 \leq x_2$ functions. Let $h:[0,1] \times [0,1] \rightarrow \mathbb{R}$ be the function $h(x,y)=f(x)g(y)$. Show h is integrable.

Theorem: Let Q be a rectangle, and let $f: Q \rightarrow \mathbb{R}$ be a bounded function. Then $\underline{\int_Q} f \leq \overline{\int_Q}f$; equality holds if and only if given $\epsilon>0$, $\exists$ a corresponding partition P of Q for which $U(f,P)-L(f,P)<\epsilon$.

Lemma: Let $Q$ be a rectangle; ;et $f: Q \rightarrow \mathbb{R}$ be a bounded function. If P and P' are any two partitions of Q, then $L(f,P) \leq U(f,P')$.

Corollary: If $f,g: Q \rightarrow \mathbb{R}$ are bounded functions on a rectangle Q such that $\{x \in Q: f(x) \neq g(x) \}$is a finite set then f is integrable if and only if g is integrable. In this case $\int_Q f=\int_Q g$.

I don't have clue so far for this question, so I am trying to list some potentially useful theorem/lemma/corollary and wonder if someone can help out. Appreciate it.

Do you know Fubini's theorem? – Coward Jul 25 '20 at 05:23 — Coward, Jul 25 '20 at 05:23
@Coward Yeah I do. I didn't think about that. – neveryield Jul 25 '20 at 17:05 — neveryield, Jul 25 '20 at 17:05

score 3 · Answer 1 · answered Jul 25 '20 at 05:24

3

Hint 1: The set of all points of discontinuity of a monotone function is countable and a countable set has measure $0$.

Hint 2: A function $f$ is integrable if and only if the set of points of discontinuity of $f$ has measure $0$

answered Jul 25 '20 at 05:24

Charlotte

1,674

1

If the hints don't serve your purpose, lemme know. I shall give a complete answer – Charlotte Jul 25 '20 at 05:28
That's helpful! – neveryield Jul 25 '20 at 17:05

score 3 · Accepted Answer · answered Jul 25 '20 at 06:06

We have $|f(x)| \leqslant M_f$ and $|g(y)| \leqslant M_g$ for all $x,y \in [0,1]$. Assume $M_f, M_g > 0$. (Otherwise we have the trivial case $h = 0$).

Since $f$ and $g$ are each bounded and monotone and, hence, Riemann integrable on $[0,1]$, there exist partitions $P' = (x_0,x_1,\ldots, x_n)$ and $P''= (y_0,y_1,\ldots, y_m)$ such that

$$U(P',f) - L(P',f) < \frac{\epsilon}{2M_g}, \quad U(P'',g) - L(P'',g) < \frac{\epsilon}{2M_f}$$

Leaving some steps for you to complete ...

(1) Forming the partition $P = (P',P,'')$ of $[0,1]^2$ show that

$$U(P,h) - L(P,h)= [U(P',f)- L(P',f)] U(P'',g) + [U(P'',g)- L(P'',g)] L(P',f)$$

(2) Show that $|U(P'',g)| \leqslant M_g$ and $|L(P',f)| \leqslant M_f$, and

$$U(P,h) - L(P,h) < \epsilon$$

Let $h:[0,1] \times [0,1] \rightarrow \mathbb{R}$ be the function $h(x,y)=f(x)g(y)$. Show h is integrable.

2 Answers2