Let $f \in L^1(X)$ where $\mu(X)<\infty$ then $g(t)=\int_X\cos(tf(x))dx$ is in $C^1(\mathbb{R})$

Question

Let $f \in L^1(X)$ where $\mu(X)<\infty$ then $g(t)=\int_X\cos(tf(x))dx$ is in $C^1(\mathbb{R})$

I have proven that the derivative is $g'(t)=\int_X-f(x)\sin(tf(x))$ using MVT and then DCT similarily to the problem I solved here: Show the function $f(t)=\int_E\sin(tx)$ is continuous When $E$ is finite measure and differentiable when $E$ is bounded. But i cannot seem to show it is continuous. I cannot use the trick as in the previous question as $X$ is not $R$ and I cannot get away with taking bounded intervals. I have no control over how $f(x)$ behaves.

Answer 1

Hint: Let $t_n \to t$. Put $h_n(x)=-f(x)\sin(t_nf(x))$. Note that $|h_n|\le |f|$. Also, if $h(x):=-f(x)\sin (tf(x))$, then $h_n \to h$ pointwise. What does DCT say now?