Is a Function of Bounded Variation Always Bounded and Integrable?

[kfdleb]

Homework Statement

f is of bounded variation on [a;b] if there exist a number K such that

$$\sum$$$$^{n}_{k=1}$$|f(a_k)-f(a_k-1)| $$\leq$$K

a=a_0<a_1<...<a_n=b; the smallest K is the total variation of f

I need to prove that
1) if f is of bounded variation on [a;b] then it is bounded on [a;b]
2) if f is of bounded variation on [a;b], then it is integrable on [a;b]

2. The attempt at a solution

i thought of using triangle inequation such that
0<=|f(b)-f(a)|<=$$\sum$$$$^{n}_{k=1}$$|f(a_k)-f(a_k-1)| $$\leq$$K

but I am not really sure how to prove the two statements
any help is really appreciated
thanks

 

[mighty2000]

Hello,

To prove that f is bounded on [a;b], we can use the fact that f is of bounded variation on [a;b]. By definition, this means that there exists a number K such that \sum^{n}_{k=1}|f(ak)-f(ak-1)| \leqK for any partition a=a_0<a_1<...<a_n=b.

Now, let's consider the partition where a=a_0=a_n and a_i=a for all i=1,2,...,n-1. In this case, we have \sum^{n}_{k=1}|f(ak)-f(ak-1)|=|f(b)-f(a)|\leqK. This shows that f is bounded on [a;b] since K is a fixed number.

To prove that f is integrable on [a;b], we can use a similar approach. We know that f is of bounded variation on [a;b], so there exists a number K such that \sum^{n}_{k=1}|f(ak)-f(ak-1)| \leqK for any partition a=a_0<a_1<...<a_n=b.

Now, let's consider the partition where a=a_0=a_n and a_i=a for all i=1,2,...,n-1. In this case, we have \sum^{n}_{k=1}|f(ak)-f(ak-1)|=|f(b)-f(a)|\leqK. This shows that f is integrable on [a;b] since the Riemann sum can be made arbitrarily small by choosing a partition where |f(b)-f(a)| is small.

I hope this helps! Let me know if you have any further questions.