Injectivity of a Fredholm operator While doing my study on the boundary-crossing time of a  stochastic process, I happened to deal with the following question which is somehow related to Fredholm theory.
Question :  Suppose $K$ is continuous non-negative function on $[0,T]^2$ such that $K(t,s)= 0 $ if and only if $s \ge t$. Prove or disprove the following statement:
"If $f$ is integrable function on $[0,T]$, then $ \int_{[0,T]} K(t,s)f(s)ds= 0  \hspace{3mm} \forall t \iff f= 0 $ a.s"
Thought : This statement is reasonable because for each $t$, $K(t, \cdot)$ can mimic the role of the indicator function $\mathbb{1}_{[0,t]}$ . 
I'm essentially looking for an answer helping me proceed with that question, for example, an approach, or a condition to be added in order to make the question more agreeable,etc. Nevertheless, no matter how your help is, I always deeply appreciate it.
Thank you.
 A: Surprisingly (to me), the statement is false. My counterexample is a little messy, but the idea is fairly simple.
Take $T = 1$ and set $a_n = \frac{1}{n}$ and $b_n = 1 - \frac{1}{n}$ for $n \in \mathbb{N}$. Define $f$ by setting $f(t) = 1$ on the intervals $[a_{2n+2}, a_{2n+1})$ and $f(t) = -1$ on the intervals $[a_{2n+1},a_{2n})$. The point of using $\frac{1}{n}$ is that the lengths of adjacent intervals are approximately equal for large $n$ (i.e., their ratio goes to $1$). We seek a family of functions $k_t(s)$ which is continuous in both $s$ and $t$, which satisfies $k_t^{-1}(\{0\}) = [t, 1]$, and such that $\int k_t(s)f(s) = 0$ for all $t$.
First we construct $k_1$. We impose the condition that $k_1(a_{2n}) = b_{2n}$ for all $n$. Then we define $k_1$ on $[a_{2n+2}, a_{2n}]$ to be any continuous function which satisfies $\int_{a_{2n+2}}^{a_{2n}} k_1f = 0$ (i.e., $\int_{a_{2n+2}}^{a_{2n+1}} k_1 = \int_{a_{2n+1}}^{a_{2n}} k_1$) and satisfies $b_{2n} \leq k_1 \leq \alpha_nb_{2n}$ where $\alpha_n \to 1$ as $n \to \infty$. (Possible because successive intervals have approximately the same length.) Then $k_1$ is continuous, its integral against $f$ is zero, and $k_1^{-1}(\{0\}) = \{1\}$.
Next, for each $n \geq 1$ let $k_{a_{2n}}(s) = k_1(s) - b_{2n}$ for $s \in [0,a_{2n}]$ and let it be $0$ for $s > a_{2n}$. These functions also have the desired properties.
To complete the construction I have to show that you can continuously interpolate $k_{a_{2n}}$ and $k_{a_{2n + 2}}$ while preserving the desired properties. You can do this by letting $k_t(s) = k_1(s) - (1-t)$ for $s \in [0,a_{2n+4}]$ (this ensures that $\int_0^{a_{2n+4}} k_t f = 0$) and dealing with the $s \in [a_{2n+4},a_{2n}]$ case separately. On the latter interval I don't have a simple formula, but I think it's clear that this can be done. If I had to write down an explicit formula I'd make each $k_t$ piecewise linear so you're just varying a couple of values of $k_t$ within $[a_{2n+4},a_{2n}]$ in a way that keeps the integral against $f$ zero and makes $k_t$ to go zero at $s = t$.
Well, that's a sketch of a counterexample.
tl;dr: for a counterexample to work, $f$ must take both positive and negative values in any neighborhood of $0$. Once such an $f$ is chosen, find functions $k_{1/n}$ which are continuous, nonzero on $[0,1/n)$ and zero on $[1/n,1]$ and whose integral against $f$ is zero. Finally, continuously interpolate between $k_{1/n}$ and $k_{1/(n+1)}$ to get functions $k_t$ with the same properties. You have to be a little careful in the last step, but there's no serious obstruction.
