Lower integral on two functions

**serious331** · May 3rd 2010, 07:11 PM

Suppose that two bounded functions $f : [a,b] \rightarrow R$ and $g : [a,b] \rightarrow R$ have the property that $g(x) \leq f(x)$ for all x in [a,b].

For a partition P of [a,b] show that $L(g,P) \leq L(f,P)$

I tried to do this by considering a partition $P = \{{x_0},...{x_n}\}$ , and for each index $i \geq 1$ , the number m is a lower bound for both the functions given x in $[{x_i-1}, {x_i}]$ . so, $m \leq m_i$
is this the correct path? I am stuck on this step..

**Drexel28** · May 3rd 2010, 07:12 PM

Originally Posted by serious331

Suppose that two bounded functions $f : [a,b] \rightarrow R$ and $g : [a,b] \rightarrow R$ have the property that $g(x) \leq f(x)$ for all x in [a,b].

For a partition P of [a,b] show that $L(g,P) \leq L(f,P)$

I tried to do this by considering a partition $P = \{{x_0},...{x_n}\}$ , and for each index $i \geq 1$ , the number m is a lower bound for both the functions given x in $[{x_i-1}, {x_i}]$ . so, $m \leq m_i$
is this the correct path? I am stuck on this step..

What are you having trouble with, precisely?

**serious331** · May 3rd 2010, 07:19 PM

Originally Posted by Drexel28

What are you having trouble with, precisely?

What I am unsure about is that if we are considering the same partition, as defined above, for both the functions, then for both the functions we end up as:

$m(b-a) = \sum_{i=1}^n m({x_i} - {x_{i-1}}) \leq \sum_{i=1}^n {m_i}({x_i} - {x_{i-1}}) \leq L(g,P)$

and $\sum_{i=1}^n {m_i}({x_i} - {x_{i-1}}) \leq L(f,P)$

Is that what I need to do?

**Drexel28** · May 3rd 2010, 07:21 PM

Originally Posted by serious331

What I am unsure about is that if we are considering the same partition, as defined above, for both the functions, then for both the functions we end up as:

$m(b-a) = \sum_{i=1}^n m({x_i} - {x_{i-1}}) \leq \sum_{i=1}^n {m_i}({x_i} - {x_{i-1}}) \leq L(g,P)$

and $\sum_{i=1}^n {m_i}({x_i} - {x_{i-1}}) \leq L(f,P)$

Is that what I need to do?

It seems like you're all messed up.

Would you agree that $f\leqslant g\implies \inf_{x\in I}f(x)\leqslant\inf_{x\in I}g(x)$ for any $I\subseteq [a,b]$ ?

**serious331** · May 3rd 2010, 07:27 PM

Originally Posted by Drexel28

It seems like you're all messed up.

Would you agree that $f\leqslant g\implies \inf_{x\in I}f(x)\leqslant\inf_{x\in I}g(x)$ for any $I\subseteq [a,b]$ ?

you're right Drexel28. I am totally confused about this chapter. Still trying to learn.

I think I would agree to the inequality you have presented, but I dont think I wont be able to write a proof on how that is true!

isnt it ${m_{i_g}} \leq {m_{i_f}}$ ?

**maddas** · May 3rd 2010, 08:03 PM

If $g\le f$ , you have that for any $x\in I$ , $\inf_{y\in I} g(y)\le g(x) \le f(x)$ , so $\inf_{y\in I} g(y)$ is a lower bound for $f(y)$ ( $y\in I$ ). Since $\inf_{y\in I} f(y)$ is the greatest lower bound of the f(y), it follows that it is $\ge$ the lower bound $\inf_{y\in I} g(y)$ .

**harish21** · May 5th 2010, 12:57 AM

Originally Posted by serious331

Suppose that two bounded functions $f : [a,b] \rightarrow R$ and $g : [a,b] \rightarrow R$ have the property that $g(x) \leq f(x)$ for all x in [a,b].

For a partition P of [a,b] show that $L(g,P) \leq L(f,P)$

I tried to do this by considering a partition $P = \{{x_0},...{x_n}\}$ , and for each index $i \geq 1$ , the number m is a lower bound for both the functions given x in $[{x_i-1}, {x_i}]$ . so, $m \leq m_i$
is this the correct path? I am stuck on this step..

Hi serious, how about this?

For any partition interval $[x_{i-1}, {x_i}]$ :

$m_{g_i} = \mbox{inf} \{g(x) : x \in [{x_{i-1}},{x_i}]\} \leq m_{f_i} = inf \{f(x) : x \in [{x_{i-1}},{x_i}]\}$

So, for each $x \in [{x_{i-1}},{x_i}]$ , we have $g(x) \leq <br /> f(x)$ . Suppose this is false & ${m_{g_i}} > {m_{f_i}}$

Let $\epsilon = \frac{{m_{g_i}} - {m_{f_i}}}{2}$ . Then, there exists $x \in [{x_{i-1}},{x_i}]$ and f(x) such that $f(x) - {m_{f_i}} < \epsilon \implies f(x) < \epsilon + {m_{f_i}} = {m_{f_i}}+ \frac{{m_{g_i}} - {m_{f_i}}}{2} = \frac{{m_{g_i}} + {m_{f_i}}}{2} < {m_{g_i}}$ which is a contradiction!

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