If the operator $T$ is defined by $Tf(x)=\int_0^xf(t)\,dt$, show that $Tf \in C[0,1]$ 
Consider the operator $T$ on $L^2[0,1]$ defined by $Tf(x)=\displaystyle \int_0^xf(t)\,dt.$ Show that $Tf \in C[0,1].$ 

I have one question before this:
What are the implications between $L^p$ spaces, i.e if $f \in L^p$ does this imply that $f \in L^{p+1}$?
My attempt:
Assume $Tf$ is not in $C[0,1]$, so there exists $y \in [0,1]$ and $\epsilon>0$ such that for all $\delta >0$, we can find $x_0$, with $|x-x_0|<\delta$ but 
$$\bigg\rVert\int_0^x f-\int_0^{x_0} f\,\bigg\rVert>\epsilon$$
WLOG assume $x>x_0$, so $$\bigg\lVert \int_{x_0}^x f\,\bigg\rVert \geq \epsilon.$$
So for $\delta_n=\frac{1}{n},$ we can find $x_n \in [0,1]$ such that $|x-x_0|<\frac{1}{n}$ and $$\bigg\lVert\int_{x_n}^xf\,\bigg\rVert \geq \epsilon$$
I don't know if that will lead me to a contradiction.
I would appreciate any help or hints with that.

Let $x_0 \in [0,1]$ and $x_n \to x_0$
Then $f_n:=f(x)1_{[0,x_n]}(x) \to  f(x)1_{[0,x_0]}(x)$
Also $|f_n| \leq |f| \in L^2$ 
So by dominated convergence we have that $Tf(x_n) \to Tf(x)$
