Let $Tf(x)=\int_{[0,x]}fdm$ for $0 \leq x \leq 1$. Prove that $T$ maps $L^1[0,1]$ into $C_0((0,1])$ and that $T$ is 1-1 Let $Tf(x)=\int_{[0,x]}fdm$ for $0 \leq x \leq 1$.
Prove that $T$ maps $L^1[0,1]$ into $C_0((0,1])$ and that $T$ is injective.

My attempt:
It is clear to me that $Tf(0)=0$. Thus I believe that to obtain continuity the dominated convergence theorem could be applied. To show that $T$ is injective, I think it is important that if $Tf(x)=0$ then $\int_{(a,b)} f(t)dt=0$ for any $a,b \in [0,1]$ with $a<b$.
I would appreciate some help with making these ideas more rigorous.. Thanks!!
 A: You can obtain continuity via Dominated Convergence: if $x_n\to x$, then $1_{[x_n,x]}\to 1_{\{x\}}$ pointwise. Then
$$
Tf(x_n)-Tf(x)=\int_{[0,1]} 1_{[x_n,x]}\,f\,dm\to0.
$$
And then you need to use that, on $[0,1]$ a function is continuous if and only if it is sequentially continuous. 
Here is a second way: assume $x<y$; and assume that $f$ is bounded, $|f|\leq c$. Then
\begin{align}
|Tf(y)-Tf(x)|=\left|\int_x^y f\,dm\right|\leq\int_x^y |f|\,dm\leq c|y-x|.
\end{align}
So $Tf$ is Lipchitz, and in particular continuous. Now fix $\varepsilon>0$. For arbitrary $f\in L^1$, there exists $g\in L^1$, bounded, with $\|g-f\|_1<\varepsilon/3$. Then
\begin{align}
|Tf(y)-Tf(x)|&\leq|Tf(y)-Tg(y)|+|Tg(y)-Tg(x)|+|Tg(x)-Tf(x)|\\[0.3cm]
&\leq\|f-g\|_1+c|y-x|+\|f-g\|_1
\end{align}
So if $|y-x|<\varepsilon/3c$, we get that $\|Tf(y)-Tf(x)|<\varepsilon$, and so $Tf$ is continuous. 
For injectivity, once you know that $\int_{[a,b]}f=0$ for all $a,b$, you can use Lebesgue Differentiation to obtain that $f=0$ a.e.
