Help with $\mathop {\lim }\limits_{a \to {0^ + }} a\int_0^1 {{x^{a - 1}}f\left( x \right)dx} = f\left( 0 \right)$ The question is as the following, where the integration is Lebesgue integration.

Let $f$ be a continuous function in the closed interval $[0,1]$. Prove that  $\mathop {\lim }\limits_{a \to {0^ + }} a\int_0^1 {{x^{a - 1}}f\left( x \right)dx}  = f\left( 0 \right)$.

My attempt is integration by parts. $a\int_0^1 {{x^{a - 1}}f\left( x \right)dx}  = a\int_0^1 {f\left( x \right)d\frac{{{x^a}}}{a}}  = \left[ {f\left( x \right){x^a}} \right]_0^1 - \int_0^1 {{x^a}df}  = f\left( 1 \right) - \int_0^1 {{x^a}df} $.
Now to prove the claim, we only need to show $f\left( 1 \right) - f(0) = \mathop {\lim }\limits_{a \to {0^ + }} \int_0^1 {{x^a}df} $, which holds if the limit and integration is exchangeable for the right-hand side. But I am not certain if such exchange is valid because it is $df$ instead of $dx$ and $f$ is not said to be differentiable on $[0,1]$.
Thank you!
 A: HINT:
Split the integral as
$$\int_0^1x^{a-1}f(x)dx=\int_0^{\delta}x^{a-1}f(x)dx+\int_{\delta}^1x^{a-1}f(x)dx$$
where for a given $\epsilon>0$, $0<x<\delta \implies |f(x)-f(0)|<\epsilon$
SPOLIER ALERT:  SCROLL OVER SHADED AREA TO DISPLAY ANSWER

Since $f$ is continuous, then for any given $\epsilon >0$, we can find a $\delta >0$, such that $|f(x)-f(0)|<\epsilon/2$ whenever $0<x<\delta$.  For such a delta, we write the integral as $$\int_0^1x^{a-1}f(x)dx=\int_0^{\delta}x^{a-1}f(x)dx+\int_{\delta}^1x^{a-1}f(x)dx \tag 1$$Now, we write the first integral on the right-hand side of $(1)$ as $$\int_0^{\delta}x^{a-1}f(x)dx=\frac{\delta^a}{a}f(0)+\int_0^{\delta}x^{a-1}(f(x)-f(0))\,dx$$Then, we have $$\left|a\int_0^1x^{a-1}f(x)dx- f(0)\right|\le(1-\delta^a)|f(0)|+ a\,\int_0^{\delta}x^{a-1}|f(x)-f(0)|\,dx+a\,\int_{\delta}^1x^{a-1}|f(x)|\,dx \le\left|\delta^a\frac{\epsilon}{2}\right|+(||f||_{\infty}+|f(0)|)(1-\delta^a)$$Now, for the given $\epsilon$, we choose $a$ so that $\delta^a>1-\frac{\epsilon}{2(||f||_{\infty}+|f(0)|)}$ and the proof is complete.

A: An alternative approach. By the Weierstrass approximation theorem, there is some polynomial $p(x)$ such that for every $x\in[0,1]$, $\left|p(x)-f(x)\right|\leq\varepsilon $. Since:
$$ \lim_{a \to 0^+}a \int_{0}^{1} x^{a-1}\cdot x^n\,dx=\lim_{a\to 0^+}\frac{a}{n+a}=0 $$
for every $n>0$ and $\int_{0}^{1}a x^{a-1}\,dx = 1$, it follows that:
$$ \lim_{a \to 0^+} a \int_{0}^{1} x^{a-1} p(x)\,dx = p(0), $$
but since $\left|p(0)-f(0)\right|\leq\varepsilon $ and:
$$ \left|a\int_{0}^{1}x^{a-1}\left(f(x)-p(x)\right)\,dx\right|\leq\varepsilon, $$
it follows that:
$$ \lim_{a\to 0^+} a\int_{0}^{1} x^{a-1}f(x)\,dx $$
exists and it is at most $2\varepsilon$ apart from $f(0)$. Since $\varepsilon$ is arbitrary, the claim follows.
