Absolute value of a function integrable?

[tomboi03]

this is the question,
Prove that if f is continuous on (a,b] and if |f| is bounded on [a,b] then f is integrable on [a,b]. (note: it is not assumed that f is continuous at a.)

I know you have to use the upper and lower bounds to prove this statement but i don't know where to start?

Thanks,
Jonnah Song

 

[HallsofIvy]

The simplest thing to do is to construct Riemann sums with base having n intervals taking one "rectangle" with base from a to a+ 1/n. Since f is bounded and continuous on (a, b], it is integrable from a+ 1/n to b and the area of the "leftmost rectangle", from a to a+ 1/n goes to 0 as 1/n goes to 0.

By the way, this doesn't have anything at all to do with |f| being integrable. In fact, I don't know why you say "|f| bounded". That is exactly equivalent to f itself being bounded which is the important thing.

 

[tomboi03]

but, that's what it says on the question. I didn't state that.

 

[tomboi03]

My TA said " this function is integrable if and only if the upper and lower limits are the same. Function is continuous, so consider the intermediate value theorem." Do you guys have any idea what he's talking about?

I'm very confused.

 

[tomboi03]

I'm still not sure how to solve this.. because..

I tried doing this...
We set s_k= inf U(f,P)
t_k= sup L(f,P)
define step function s and t by letting their values equal
s_k and t_k on (x_k-1, x_k)
s(x_k)= t(x_k)= f(x_k)
s(x)\leqf(x)\leqt(x) for all x
t_k-s_k<E
\int(a+1/n, b) t - \int(a+1/n, b) = \int(a+1/n, b)(t-s) \leq E(b-a)
therefore f is integrable on [a+1/n, b]

On the other hand,
for [a, a+1/n]
lim \int (a, a+1/n) f\leqE(a+1/n-a)
n\rightarrow0
lim \int (a, a+1/n) f\leq E(1/n)
therefore f is integrable on [a,a+1/n]
therefore f is integrable on [a,b]

what am i doing wrong?
because i got this wrong.