Differentiation and Lebesgue integration

[AxiomOfChoice]

Homework Statement

Suppose $g(x) = \int_0^x f(t) dt$, where $f$ is Lebesgue integrable on $\mathbb R$. Give an $\epsilon - \delta$ proof that $g'(y) = f(y)$ if $y\in (0,\infty)$ is a point of continuity of $f$.

Homework Equations

The Attempt at a Solution

I know I need to show that
$$f(y) = \lim_{h\to 0} \int_y^{y+h} \frac{1}{h} f(t) dt.$$
My idea was to try to do this in terms of sequences; i.e., to let $\{h_n\}$ be any sequence of real numbers such that $h_n \to 0$, and then to phrase the limit above in terms of a limit as $n\to \infty$. I had then planned to use something like the dominated convergence theorem. But I don't have any idea how to make use of the hypothesis that $f$ is continuous at $y$, so I'm not sure if this is the right approach.

 

[lanedance]

could you look at it as follows:
$$f(y) = \lim_{h\to 0} \frac{1}{h} \int_y^{y+h} f(t) dt$$

then use the conitinuity of f to show that as h gets small
$$\int_y^{y+h} f(t) dt \approx f(t).h$$

 

[AxiomOfChoice]

Ok. I think I know how to do this. Here is a complete solution: Let $f \in L^1(\mathbb R)$. We want to show that, given any $\epsilon > 0$, we can find a $\delta > 0$ such that
$$\left| \frac{1}{h} \int_y^{y+h} f - f(y) \right| < \epsilon$$
whenever $|h| < \delta$. Since $f$ is continuous at $y$, we know we can find a $\delta$ such that $|t - y| < \delta$ implies $|f(t) - f(y)| < \epsilon$. So, if we make $|h| < \delta$, we have
$$\begin{align*}<br /> \left| \frac{1}{h} \int_y^{y+h} f - f(y) \frac{1}{h} \int_y^{y+h} f \right|<br /> & \leq \frac{1}{h} \int_y^{y+h} \left| f(t) - f(y) \right| dt\\<br /> & \leq \frac{1}{h} \int_y^{y+h} \epsilon dt = \epsilon.<br /> \end{align*}$$

That does it, I think.

 

[hunt_mat]

There is a typo, you should write:
$$\left| \frac{1}{h} \int_y^{y+h} f - f(y) \right| =\Bigg|\frac{1}{h}\int_{y}^{y+h}f-f(y)\int_{y}^{y+h}\frac{1}{h}\Bigg|$$