I know this question is very similar to many asked but I have looked and could not find an exact match.

I have an integral formula which I know to be true, for all $\epsilon \in [0,h)$ for some $h \in \mathbb{R}$:

$$\int_Kf_{\epsilon}(x)d\mu(x) = Pf_\epsilon(k)$$


  • $P$ is some constant,
  • $\{f_\epsilon\}_{\epsilon \in [0,h)}$ is some set of functions (all defined on the same domain, denote the domain $D$)
  • $k$ is some constant in $D$
  • $K$ is some (not necessarily compact) subset of $D$
  • $\mu$ is some measure.

I also know that $f_\epsilon(k)$ is continuous with respect to $\epsilon$ over $(0,h)$.

Define $\hat{f}(y)$ by; for all $y \in D$:

$$\hat{f}(y) = \lim_{\epsilon \to 0} f_\epsilon(y)$$

$\hat{f}(k)$ might not equal $f_0(k)$ but is always finite. My question is; does it follow that:

$$\int_K\hat{f}(x)d\mu(x) = P\hat{f}(k)$$

And if not, under what conditions does it fail? Would anyone be able to give me an example?

Thank you in advance!

  • 1
    $\begingroup$ I don't understand your notation, why is there a $k$ at the right-hand side of the equation. As I think from the left-hand side $k$ is the integration variable, over some (I am assuming compact) set $K$, if that's true, there shouldn't be any $k$ at the right-hand side. $\endgroup$ – David Jaramillo Aug 23 '18 at 21:47
  • $\begingroup$ Apologies yes - this should now be changed - though K is not necessarily compact... I don't think... $\endgroup$ – JDoe2 Aug 23 '18 at 22:25
  • $\begingroup$ hmmmm I think that doesn't solve the problem. Whats $k$ at the right-hand side $\endgroup$ – David Jaramillo Aug 24 '18 at 1:28
  • $\begingroup$ Sorry :/ Capital $K$ is some set over which the integral is being performed. Lower case $k$ is some constant in the domain of $f$. $\endgroup$ – JDoe2 Aug 24 '18 at 11:26
  • $\begingroup$ Your notation is impossible to understand. What is $f_{0}$, what is $x$ in your second sentence ? Is it fixed ? In such a case, you should not use the notation $x$ in the integral. Then, what is $K$ ? Finally, why do you know that the limit $\hat{f}$ exist ? $\endgroup$ – M. Dus Aug 24 '18 at 11:38

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