Proving an $L^p$ limit Let $f$ be a $L^{p}$ function on $\mathbb{R}$. If $p>4/3$, prove
that:
$$\lim_{t\rightarrow0^{+}}\int_{0}^{t}x^{-1/4}f(x)dx=0.$$
The natural procedure here is to bound the integral by using Hölder's
inequality; after using the Hölder inequality, we get that as $t\rightarrow0^{+}$
the integral from $0$ to $t$ of $\left|x^{-1/4}f(x)\right|$ goes
to zero. Does the desired result follow immediately from the Dominated
Convergence Theorem?
 A: If $pq=p+q$, then $q<4$, so $x^{-1/4}$ is in $L^q$ on $(0,t)$ for all $t>0$.  Hence you can apply Hölder's inequality as you indicated to conclude that $x^{-1/4}f(x)$ is integrable on each $(0,t)$.  
Yes, you can apply dominated convergence.  For example, the integrable function $x^{-1/4}|f(x)|\chi_{(0,1)}(x)$ dominates the functions $x^{-1/4}|f(x)|\chi_{(0,t)}(x)$ for $t<1$, which converge pointwise to $0$ as $t\to 0$.
But you could also apply the fact that if $g$ is integrable, then for all $\varepsilon>0$ there exists $\delta>0$ such that $m(A)<\delta$ implies $|\int_A g|<\varepsilon$.  In other words, you could apply absolute continuity of the integral.
A: To use the dominated convergence theorem, you have to show that the integrands $x^{-1/4}|f(x)|\chi_{(0,t)}$ are bounded by an integrable function, and this essentially means that you have to show $$\int_0 ^\varepsilon x^{-1/4} |f(x)|dx < \infty$$
for an $\varepsilon>0$ of your choice. This appears slightly weaker than your claim, but you still need Hölder's inequality to show this.
