Change order of integration to show an equality I'm trying to show the following equality using a change of integration order, but I'm unable to complete it:

$$ \int_y^1\int_0^t f(x)\,\mathrm dx\,\mathrm  dt \stackrel{?}{=}\int_0^1 k(y,x)f(x)\,\mathrm dx$$
for some $k \in C([0,1]^2)$.

where $f \in C[0,1]$.
Here's what I've done so far (I might have a mistake in the change of order limits, but the idea was to capture the fact that $t$ depends either on $x$ or $y$, depending on the which is larger):
$$ \int_y^1\int_0^t f(x)\,\mathrm dx \,\mathrm dt = \int_0^1 \int_x^1 f(t)\chi_{[y,1]}(t)\,\mathrm dt\,\mathrm dx$$
(where $\chi_{[y,1]}$ is the indicator function).
EDIT: A closer option is splitting by whether $x<y$ or not, something as such:
$$  
k(x,y) := 
     \begin{cases}
       \int_x^1 f(t)dt &\quad\text{if } 0\leq y \leq x\\
       \int_y^1 f(t)dt &\quad\text{if } x< y \leq 1 \\
     \end{cases}
$$
But I'm still missing the "$f(x)$" term on its own, as in the RHS of the equality.
This is as close as I got to showing there's some $k$ that fits the bill on the left-hand-side in the equality, but I'm not sure how to proceed.
Any ideas?
 A: Let $$f_n(x) = \begin{cases}
0,\ x < p - \frac{1}{n}\\
1,\ x > p + \frac{1}{n}\\
\frac{x - p + \frac{1}{n}}{\frac{2}{n}}
\end{cases}$$
Obviously, $f_n \to \chi_{[0, p]}$, and for both left and right side limit can be moved under integral.
So, our property should hold for $f(x) = \chi){[0, p]}$ too.
Right side becomes $\int_0^p k(y, x)\, dx$.
If $y > p$, left side becomes $p - py$. If $y \leq p$, left side becomes $(1 - p)\cdot p + \int_y^p t\, dt = p - p^2 / 2 - y^2 /2$.
So, for any $p$, we have
$$\int_0^p k(y, x)\, dx = \begin{cases}
p - p^2 / 2 - y^2 / 2,\ y \leq p\\
p - py,\ y > p
\end{cases}$$
Differentiating both sides by $p$, we get
$$k(y, x) = \begin{cases}
1 - x, y \leq x\\
1 - y, y > x
\end{cases}$$
Or, combining, $k(y, x) = 1 - \max(x, y)$ (also, you can get the same $k$ substituting $f(x, y) = 1$ in your last result).
The same reasoning shows that this $k$ works:

*

*It works for $\chi_{[0, p]}$ (by derivation)

*If the equality is true for $f$ and $g$ then it's true for $f + g$ (easy to check)

*If $f_n$ are all bounded by constant $M$, and $f_n \to f$ a.e., and the equality works for each $f_n$, then it works for $f$ (any theorem about convergence should work)

*From 1 and 2, it works for any indicator of any segment $\chi_{[q, p]}$.

*Any continuous function can be approximated by linear combination of indicators of segments.

