# Thread: Is this function integrable?

1. ## Is this function integrable?

If f is a bounded function from [a,b] into the real numbers that is
integrable on [a,b], then for any c such that a < c < b, is f
integrable on [a,c] and is f integrable on [c,b]?

I want to say no, since if I pick $f(x)= \frac {1}{x}$, then f is integrable from [-2,2] but not if you move the boundary to 0.

Is that right? Thanks.

2. I have also observed same question in my book..

3. So I guess this theorem is true then?

Here is my proof so far:

Let $\epsilon > 0$ be given. Since f is integrable, then there exists a partition $P= \{ x_0 = a, x_1, x_2 , . . . , x_{N-1}, x_N = b \}$ on [a,b] such that $U(f,P)-L(f,P)< \epsilon$

Define $M_j=sup \{ f(x) : x \in [x_j,x_{j+1} \}$ and $m_j=inf \{ f(x) : x \in [x_j,x_{j+1} \}$, so we have $U(f,P)-L(f,P)= \sum ^{N-1}_{j=0}(M_j)(x_{j+1}-x_{j}) - \sum ^{N-1}_{j=0}(m_j)(x_{j+1}-x_{j}) =$ $\sum ^{N-1}_{j=0}(M_j-m_j)(x_{j+1}-x_{j})$

Now let $c=x_k \ \ \ \ \ k \in (0,N)$, then $U(f,P)-L(f,P)= \sum ^{N-1}_{j=0}(M_j-m_j)(x_{j+1}-x_{j}) =$
$\sum ^{k}_{j=0}(M_j-m_j)(x_{j+1}-x_{j}) + \sum ^{N-1}_{j=k+1}(M_j-m_j)(x_{j+1}-x_{j}) < \epsilon$

Define $P_1 = \{ x_0, x_1 , . . . , x_k = c \}$ and $P_2 = \{ c = x_k, x_{k+1}, . . . , x_N \}$

But we know that
$\sum ^{k}_{j=0}(M_j-m_j)(x_{j+1}-x_{j}) + \sum ^{N-1}_{j=k+1}(M_j-m_j)(x_{j+1}-x_{j}) =$
$\sum ^{k}_{j=0}(M_j)(x_{j+1}-x_{j}) - \sum ^{k}_{j=0}(m_j)(x_{j+1}-x_{j}) +$ $\sum ^{N-1}_{j=k+1}(M_j)(x_{j+1}-x_{j}) - \sum ^{N-1}_{j=k+1}(m_j)(x_{j+1}-x_{j}) =$
$U(P_1,f)-L(P_1,f)+U(P_2,f)-U(P_2,f) < \epsilon$

I need to show that $U(P_1,f)-L(P_1,f) < \epsilon$ and $U(P_2,f)-U(P_2,f) < \epsilon$

Any hints? Thanks.

4. Note that the theorem says that the function is bounded.
I don’t know what definition of integrable you are using.
However, if you are using some variation of subdivisions we can always include c as an endpoint in a refinement of the subdivision.

If $f$ is integrable on $[a,b]$ then $f$ is integrable on $[c,d]\subset [a,b]$.