How to use lambda as std::unique_ptr's Deleter?

Check following contrived program:

#include <functional>
#include <memory>

template<typename T>
using UniPtr = std::unique_ptr<T, std::function<void(T*)>>;

int* alloc()
{
    return new int;
}

UniPtr<int> func()
{
    auto dealloc = [](int* p){delete p;};

    return UniPtr<int>{alloc(), dealloc};
}

int main()
{
    auto p = func();
    return 0;
}

From std::function constructor manual, I think constructing std::function object may throw exception, even the ratio is very low:

UniPtr<int> func()
{
    auto dealloc = [](int* p){delete p;};

    return UniPtr<int>{alloc(), dealloc};
}

But if using function pointer instead of std::function object:

template<typename T>
using UniPtr = std::unique_ptr<T, void(*)(T*)>;

I think after leaving the func() scope, the dealloc object should be freed, and it can't be referenced. Please correct me if I am wrong. So the only safe method I can come out is defining a global dealloc function:

void dealloc(int* p)
{
    delete p;
}

But I don't like this method.

Based on precedent exposition, there is not 100% safe way to use lambda as std::unique_ptr's Deleter, Or I misunderstand something? How to use lambda as std::unique_ptr's Deleter?