# Show that Lebesgue integral is continuous and differentiable

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be given by $$g(t)=\int_\left[0,1\right]e^\sqrt{x^2+t^2}d\lambda(x), t \in \mathbb{R}$$ a) Justify that the integral is defined for each $t \in \mathbb{R}$ and show next that g is continuous. b) Show that g is differentiable and compute $g'(t)$ for all $t \in \mathbb{R}$. Find in particular $g'(0)$

My approach for (a) was to use the continuity lemma of lebesgue integrals however I am obviously unable to find a function $w(x) \in \mathcal{L^1_+}:\left|u(t,x)\right|\leq w(x)\forall (t,x) \in \mathbb{R}\times [0,1]$

For (b) I run into the same problem since the partial derivative w.r.t. t isn't bounded either. I'm sorta suspecting that my teacher has made a mistake and forgot to include a minus before the squareroot, however I might be mistaken, thus I'm asking here.

• See Robert's answer below: If you show that, for an arbitrary $K>0$ that $g$ satisfies the conditions for $t \in (-K,K)$, then the result is true for $\mathbb{R}$. – copper.hat Nov 8 '16 at 0:26
• I've shown that g satisfies the conditions for $t \in \left[ -K,K\right]$. Is there any strict way of proving that it then applies to all of $\mathbb{R}$ or would you say it is trivial? – O. T. Nov 8 '16 at 8:19
• You are trying to prove some proposition $P(t)$ for all $t \in \mathbb{R}$. If you prove that for all $K>0$, then $P(t)$ is true for all $t \in [-K,K]$ then this implies $P(t)$ for all $t \in \mathbb{R}$. – copper.hat Nov 8 '16 at 14:35

Hint: You can't get a single $w$ that works for all $t \in \mathbb R$, but for any $K> 0$ you can get one that works for all $|t| \le K$.
• I think I can follow your reasoning. How would I then go about expanding from $\left|t\right|\leq K$ to $t \in \mathbb{R}$? :) – O. T. Nov 8 '16 at 0:13